Applied Optimization is pretty tough,
finishing the homework has been rough.
But this topic is prevalent in many fields,
so I should probably keep my dislike concealed.
Problem:
#9. Suppose 600 ft. of fencing are used to enclose a corral in the shape of a rectangle with a semi-circle whose diameter is a side of a rectangle. Find the dimension of the corral with the maximum area.
Perimenter=2x+y+(pi*y)/2
600=2x+y+(pi*y)/2
x=(600-y-(pi/2)*y)/2
A(x)=(xy+pi(y/2)^2)/2
=(600y-(y^2)+pi/2*y^2)/2 + 2+(pi*y)/8
=300y-(y^2)/2- (1/8)*pi*y^2
A'(x)= 300 - (2y/2)-(1/4)*pi*y
A'(x)= 300-y-(1/4)*pi*y =0
y=1200/(4+pi)
x=(600-y-(pi/2)*y)/2
=600/(4+pi)
So in order to get the maximum amound of space, the dimensions of the corral need to be (600/(4+pi), 1200/(4+pi))
Saturday, July 31, 2010
4.5: Graph Sketching and Asymptotes
Graph sketching depends on points of transition;
points where the sign of the slope changes position.
Problem:
1. Perform all necessary steps needed to sketch the graph of the function:
f(x)=x^3-3x^2
f'(x)=3x^2-6x =0
3x(x-2)=0
3x=0; x=0 & (x-2)=0;x=2
f(0)=0^3-3(0)^2 = 0
f(2)=2^3-3(2)^2 = -4
Critical Points: (0,0) & (2,-4)
For the sign graph of the first derivative: f'(-1)=9 - indicating a positive slope, f'(1)=-3 - indicating a negative slope and f'(3)=9 - indicating a positive slope.
f''(x)= 6x-6 =0
x=1
f(1)=1^3-3(1)^2 = -2
Point of Inflection: (1,-2)
For the sign graph of the second derivative: f''(0)=-6 - indicating a negative slope, f''(2)=6 - indicating a positive slope.
When the sign graph of the first and second derivatives are combined, the new sign graph looks like this: +- 0 -- 1 -+ 2 ++. The two zeros of the graph are located at 0 and 2 and the point of inflection, where the slope of the graph begins to change direction is at 1. Using the combined sign graphs from the first and second derivatives, I was able to draw a sketch of the graph of this function:
points where the sign of the slope changes position.
Problem:
1. Perform all necessary steps needed to sketch the graph of the function:
f(x)=x^3-3x^2
f'(x)=3x^2-6x =0
3x(x-2)=0
3x=0; x=0 & (x-2)=0;x=2
f(0)=0^3-3(0)^2 = 0
f(2)=2^3-3(2)^2 = -4
Critical Points: (0,0) & (2,-4)
For the sign graph of the first derivative: f'(-1)=9 - indicating a positive slope, f'(1)=-3 - indicating a negative slope and f'(3)=9 - indicating a positive slope.
f''(x)= 6x-6 =0
x=1
f(1)=1^3-3(1)^2 = -2
Point of Inflection: (1,-2)
For the sign graph of the second derivative: f''(0)=-6 - indicating a negative slope, f''(2)=6 - indicating a positive slope.
When the sign graph of the first and second derivatives are combined, the new sign graph looks like this: +- 0 -- 1 -+ 2 ++. The two zeros of the graph are located at 0 and 2 and the point of inflection, where the slope of the graph begins to change direction is at 1. Using the combined sign graphs from the first and second derivatives, I was able to draw a sketch of the graph of this function:
4.4: The Shape of a Graph
The shape of the graph is either up or down,
the shape of a smile or the shape of a frown.
The graph will contain a few points of inflection,
it's really easy to do if you follow directions.
To find the shape use the second derivative test,
Using a sign chart will end your quest.
Problem:
1. Using the equation f(x)=x^4-4x^3, find the second derivative, the critical points at the second derivative and where the shape of the graph is concave up/down.
f(x)=x^4-4x^3
f'(x)=4x^3-12x^2
f''(x)=12x^2-24x
f''(x)=12x^2-24x=0
12x(x-2)=0
12x=0; x=0
x-2=0; x=2
f(0)=0^4-4(0)^3=0. The first critical point is (0,0)
f(2)= 2^4-4(2)^3=-16. The second critical point is (2,-16)
To determine the sign of the graph before and after each critical point, I picked the numbers -1, 1, and 3. At f''(-1) the slope of the line is positive, at f''(1), the slope of the line is negative and at f''(3) the slope of the line is positive. This indicates that the shape of the graph at -1 is concave down and at 1 it is concave up.
the shape of a smile or the shape of a frown.
The graph will contain a few points of inflection,
it's really easy to do if you follow directions.
To find the shape use the second derivative test,
Using a sign chart will end your quest.
Problem:
1. Using the equation f(x)=x^4-4x^3, find the second derivative, the critical points at the second derivative and where the shape of the graph is concave up/down.
f(x)=x^4-4x^3
f'(x)=4x^3-12x^2
f''(x)=12x^2-24x
f''(x)=12x^2-24x=0
12x(x-2)=0
12x=0; x=0
x-2=0; x=2
f(0)=0^4-4(0)^3=0. The first critical point is (0,0)
f(2)= 2^4-4(2)^3=-16. The second critical point is (2,-16)
To determine the sign of the graph before and after each critical point, I picked the numbers -1, 1, and 3. At f''(-1) the slope of the line is positive, at f''(1), the slope of the line is negative and at f''(3) the slope of the line is positive. This indicates that the shape of the graph at -1 is concave down and at 1 it is concave up.
Friday, July 30, 2010
4.3: The Mean Value Theorem and Monotonicity
Use the derivative to test the sign,
which will help determine the graph's design.
The change in sign will determine a local max or local min,
finally figuring this out made me grin!
Problem:
1. Find the critical point(s) and apply the first derivative test. Is there a local max or a local min at the critical point(s)?
f(x)=x^2+6x-7
f'(x)=2x+6=0
f'(x)=2(x+3)=0
2 can't be equal to zero.
x+3=0, x=-3.
F(x)=(-3)^2+6(-3)-7
f(x)=9+(-18)-7=-16
Critical point: (-3,-16)
I picked points to the left and right of -3. I used -4 and -2. I plugged -4 in for x in the derivative of the original function, and got -2. I plugged -2 in for x in the derivative of the original function and got 2. This means that there is a local minimum at -3, because the sign changes from - to +.
which will help determine the graph's design.
The change in sign will determine a local max or local min,
finally figuring this out made me grin!
Problem:
1. Find the critical point(s) and apply the first derivative test. Is there a local max or a local min at the critical point(s)?
f(x)=x^2+6x-7
f'(x)=2x+6=0
f'(x)=2(x+3)=0
2 can't be equal to zero.
x+3=0, x=-3.
F(x)=(-3)^2+6(-3)-7
f(x)=9+(-18)-7=-16
Critical point: (-3,-16)
I picked points to the left and right of -3. I used -4 and -2. I plugged -4 in for x in the derivative of the original function, and got -2. I plugged -2 in for x in the derivative of the original function and got 2. This means that there is a local minimum at -3, because the sign changes from - to +.
4.2: Extreme Values
Finding the derivative is one of my favorite parts,
they're used to find extrema on various charts.
The critical points are easy to find,
they're the horizontal tangent line.
Problem:
1. Find the critical point(s) at f(x)=9x^2-5x+4.
f'(x)=18x-5
f'(x)=18x-5=0
f'(x)=18x=5
x=5/18.
f(x)=9(5/18)^2-5(5/18)+4 = 3.305
y=3.305
The critical point for this function is ((5/18),3.305)
they're used to find extrema on various charts.
The critical points are easy to find,
they're the horizontal tangent line.
Problem:
1. Find the critical point(s) at f(x)=9x^2-5x+4.
f'(x)=18x-5
f'(x)=18x-5=0
f'(x)=18x=5
x=5/18.
f(x)=9(5/18)^2-5(5/18)+4 = 3.305
y=3.305
The critical point for this function is ((5/18),3.305)
Thursday, July 29, 2010
4.1: Linear Approximation and Applications
Linear Approximation is easy to understand,
but I still find calculus to be fairly bland.
I prefer Approximation to Linearization,
but I'd still rather enjoy summer vacation.
Problem:
Find the linearization of f(x)=cos(x) + 3x^(5/2), when x=11. Then find the linear approximation at x=12.
f'(x)=-sin(x)+ (15/2)x^(3/2)
f'(11)=-sin(11) + (15/2)x^(3/2)=274.61
f(11)= cos(11) + 3(11)^(5/2)=12.03.94
L(x)=f'(11)(x-11)+f(11)
at x=12...
f(12)=f'(11)(12-11)+f(11)
f(12)=(274.61)(1)+(1203.94)
f(12)=1478.55
but I still find calculus to be fairly bland.
I prefer Approximation to Linearization,
but I'd still rather enjoy summer vacation.
Problem:
Find the linearization of f(x)=cos(x) + 3x^(5/2), when x=11. Then find the linear approximation at x=12.
f'(x)=-sin(x)+ (15/2)x^(3/2)
f'(11)=-sin(11) + (15/2)x^(3/2)=274.61
f(11)= cos(11) + 3(11)^(5/2)=12.03.94
L(x)=f'(11)(x-11)+f(11)
at x=12...
f(12)=f'(11)(12-11)+f(11)
f(12)=(274.61)(1)+(1203.94)
f(12)=1478.55
Sunday, July 25, 2010
3.11: Related Rates
Related rates problems are really interesting. It's pretty cool how you can actually use derivatives to find out something that is related to real life. I know that I am definitely one of those obnoxious, pessimistic people that always complains about math, saying that I'll never use the stuff I learn again in my life, so for me its actually really cool to see how derivatives and rates of change are used in real life situations..
Problem:
Real life example: Every summer I visit my dad in New Jersey, last year when I went to visit my dad my sister, Taryn, flew from New Jersey to the Florida Keys to visit our cousins. If I am on a plane approximately 500 miles west of Newark Airport traveling 250 miles per hour and Taryn is on a plane 50 miles south of Newark airport traveling 210 miles per hour, how fast is the distance between us changing?
a^2+b^2=c^2
a=500
b=50
c=?
500^2+50^2=c^2
250,000 + 2500=252500
c= sq. root of 252500 = 502.49
da/dt=250 mph
db/dt=175 mph
dc/dt=?
d/dt(a^2+b^2)=d/dt c^2
2a*da/dt + 2b*db/dt = 2c*dc/dt
-(2*500*250) + (2*50*175) = (2*502.49*dc/dt)
-250000+17500=1004.98*dc/dt
-232,500/1004.98=dc/dt
-231.34 mph = dc/dt
Problem:
Real life example: Every summer I visit my dad in New Jersey, last year when I went to visit my dad my sister, Taryn, flew from New Jersey to the Florida Keys to visit our cousins. If I am on a plane approximately 500 miles west of Newark Airport traveling 250 miles per hour and Taryn is on a plane 50 miles south of Newark airport traveling 210 miles per hour, how fast is the distance between us changing?
a^2+b^2=c^2
a=500
b=50
c=?
500^2+50^2=c^2
250,000 + 2500=252500
c= sq. root of 252500 = 502.49
da/dt=250 mph
db/dt=175 mph
dc/dt=?
d/dt(a^2+b^2)=d/dt c^2
2a*da/dt + 2b*db/dt = 2c*dc/dt
-(2*500*250) + (2*50*175) = (2*502.49*dc/dt)
-250000+17500=1004.98*dc/dt
-232,500/1004.98=dc/dt
-231.34 mph = dc/dt
Thursday, July 22, 2010
3.10: Derivatives of General Exponential and Logarithmic Functions
Exponential and Logarithmic derivatives are super easy. Making up derivative problems has also gotten easier.. I feel like I could sit here all day making up problems and solving.. I'm actually having fun..
Problem:
1.Find the derivative of csc(ln15x)*(11^x^2 + tan(x^(4/5))).
Use the product rule: csc(ln15x)*(11^x^2 + tan(x^(4/5)))' + (csc(ln15x))'*(11^x^2 + tan(x^(4/5)))
csc(ln15x)*(ln 11*11^x)(2x) + sec^2(x^(4/5))((4/5)x^(-1/5)) - csc(ln15x)*cot(ln15x*(1/15x)*(15)*(11^x^2 + tan(x^(4/5)))
Problem:
1.Find the derivative of csc(ln15x)*(11^x^2 + tan(x^(4/5))).
Use the product rule: csc(ln15x)*(11^x^2 + tan(x^(4/5)))' + (csc(ln15x))'*(11^x^2 + tan(x^(4/5)))
csc(ln15x)*(ln 11*11^x)(2x) + sec^2(x^(4/5))((4/5)x^(-1/5)) - csc(ln15x)*cot(ln15x*(1/15x)*(15)*(11^x^2 + tan(x^(4/5)))
3.8:Implicit Differentiation
I think implicit differentiation is pretty difficult. I think having two different variables and needing to know what to do with each can be hard sometimes.. I am not a fan of this either.
Problem:
1. Find dy/dx: 4xy - y^4 = 5x
First take the derivative of each side: d/dx(4xy-y^4) = d/dx(5x)
Then, for the left side of the equation I used the sum difference rule and took d/dx(4xy) and d/dx(y^4): d/dx(4xy)=4x*dy/dx + 4y and d/dx(y^4) =4y^3
and d/dx(5x)=5. So, 4x*dy/dx + 4y - 4y^3 = 5. To work on getting dy/dx by itself I subtracted 4y from the left side of the equation and moved it to the right side, I also added 4y^3 on the left to move it over to the right. 4x*dy/dx =5-4y + 4y^3. Then I divided by 4x. So dy/dx=(5-4y+4y^3)/(4x)
Problem:
1. Find dy/dx: 4xy - y^4 = 5x
First take the derivative of each side: d/dx(4xy-y^4) = d/dx(5x)
Then, for the left side of the equation I used the sum difference rule and took d/dx(4xy) and d/dx(y^4): d/dx(4xy)=4x*dy/dx + 4y and d/dx(y^4) =4y^3
and d/dx(5x)=5. So, 4x*dy/dx + 4y - 4y^3 = 5. To work on getting dy/dx by itself I subtracted 4y from the left side of the equation and moved it to the right side, I also added 4y^3 on the left to move it over to the right. 4x*dy/dx =5-4y + 4y^3. Then I divided by 4x. So dy/dx=(5-4y+4y^3)/(4x)
3.7: The Chain Rule
I am not a fan of the chain rule. It's super confusing when there's a lot going on in the function. The double chain rule is even worse..
Problem:
1. Find the derivative of: (x^3 + sec(ln5x) + sin(4^x) + e^x + 9x) / (cos(4x* 4th root of x))
First you use the quotient rule: (cos(4x*4th root of x)) * (x^3 + sec(ln5x) + sin(4x) + e^x + 9x)' - (x^3 + sec(ln5x) + sin(4^x) + e^x +9x) * (cos 4x*4th root of x)' ALL divided by (cos(4x*4th root of x))^2
Using a whole combination of rules, including the chain rule, the derivative of this function is:
cos(4x*4th root of x) * (3x^2) + sec(ln5x)*tan(ln5x)*(1/5x)*(5) + cos(4x)(ln4*4^x)+(e^x)+(9) - (x^3) + sec(ln5x)) + sin(4^x) + (e^x) +(9x) (-sin(4x*4th root of x)* (4)*((1/4)x^(-3/4)) ALL divided by (cos(4x*4th root of x))^2
I hope you can follow this, it looks crazy..
Problem:
1. Find the derivative of: (x^3 + sec(ln5x) + sin(4^x) + e^x + 9x) / (cos(4x* 4th root of x))
First you use the quotient rule: (cos(4x*4th root of x)) * (x^3 + sec(ln5x) + sin(4x) + e^x + 9x)' - (x^3 + sec(ln5x) + sin(4^x) + e^x +9x) * (cos 4x*4th root of x)' ALL divided by (cos(4x*4th root of x))^2
Using a whole combination of rules, including the chain rule, the derivative of this function is:
cos(4x*4th root of x) * (3x^2) + sec(ln5x)*tan(ln5x)*(1/5x)*(5) + cos(4x)(ln4*4^x)+(e^x)+(9) - (x^3) + sec(ln5x)) + sin(4^x) + (e^x) +(9x) (-sin(4x*4th root of x)* (4)*((1/4)x^(-3/4)) ALL divided by (cos(4x*4th root of x))^2
I hope you can follow this, it looks crazy..
3.6: Trigonometric Functions
This section is just ridiculous. A whole section devoted to telling us that the derivative of sin x = cos x, etc. etc. etc. This should have been stuck in with some other section.....
Problem:
1. Find the derivative of f(x): (cot(2x+4))*(sec(cubed root of x))
First you use the product rule: (cot(2x+4))'(sec(cubed root of x)) + (cot(2x+4))(sec(cubed root of x))'
Then you use the chain rule to solve: (-csc^2(2x+4))*(2)*(sec(cubed root of x) + (cot(2x+4))*(sec(cubed root of x))*(tan(cubed root of x))*((1/3)x^(-2/3))
Problem:
1. Find the derivative of f(x): (cot(2x+4))*(sec(cubed root of x))
First you use the product rule: (cot(2x+4))'(sec(cubed root of x)) + (cot(2x+4))(sec(cubed root of x))'
Then you use the chain rule to solve: (-csc^2(2x+4))*(2)*(sec(cubed root of x) + (cot(2x+4))*(sec(cubed root of x))*(tan(cubed root of x))*((1/3)x^(-2/3))
3.5: Higher Derivatives
Higher derivatives are awesome! They're so simple! Although, I would probably think they were difficult if I didn't know how to derivatives at all..
Problem:
1. Find f(4)(x) of: x^9-5x^5-cos(x^3)+7x
f'(x): 9x^8 - 25x^4 + sin(x^3)(3x^2) + 7
f''(x): 72x^7 - 100x^3 + cos(x^3)(6x) + 0
f'''(x): 504x^6 - 300x^2 - sin(x^3)(6) + 0
f(4)(x): 3,024x^5 - 600x - cos(x^3)(0) + 0
Problem:
1. Find f(4)(x) of: x^9-5x^5-cos(x^3)+7x
f'(x): 9x^8 - 25x^4 + sin(x^3)(3x^2) + 7
f''(x): 72x^7 - 100x^3 + cos(x^3)(6x) + 0
f'''(x): 504x^6 - 300x^2 - sin(x^3)(6) + 0
f(4)(x): 3,024x^5 - 600x - cos(x^3)(0) + 0
Wednesday, July 21, 2010
3.4: Rates of Change
When I took physics we used rate of change all the time for velocity and acceleration and I could never understand what was going on, now I completely understand..
Problem
1. A rock is thrown off a 550m cliff, what is its velocity when it hits the ground?
I used the formula S(t)=s0+v0t-1/2gt^2 where s0= original position (550m) v0=velocity at t (0 m/s) and g=the acceleration due to gravity (9.8m/s^2), and the formula v(t)=v0-g(t).
s(t)=550-4.9t^2 (the 4.9 came from taking 1/2 of 9.8.) Because the question asks for the velocity when the rock hits the ground, the height is equal to 0. So I set the equation for s(t) equal to 0.
0=550-4.9t^2, solving for t I got t=square root of (550/4.9).
Since the v(t)=v0-g(t), v(t)=-9.8t
To find the velocity I plugged in the square root of (550/4.9) for t and multiplied it by -9.8. The velocity of the rock when it hits the ground is equal to -103.82 m/s. The number is negative because the rock is falling, therefore decreasing in velocity.
Problem
1. A rock is thrown off a 550m cliff, what is its velocity when it hits the ground?
I used the formula S(t)=s0+v0t-1/2gt^2 where s0= original position (550m) v0=velocity at t (0 m/s) and g=the acceleration due to gravity (9.8m/s^2), and the formula v(t)=v0-g(t).
s(t)=550-4.9t^2 (the 4.9 came from taking 1/2 of 9.8.) Because the question asks for the velocity when the rock hits the ground, the height is equal to 0. So I set the equation for s(t) equal to 0.
0=550-4.9t^2, solving for t I got t=square root of (550/4.9).
Since the v(t)=v0-g(t), v(t)=-9.8t
To find the velocity I plugged in the square root of (550/4.9) for t and multiplied it by -9.8. The velocity of the rock when it hits the ground is equal to -103.82 m/s. The number is negative because the rock is falling, therefore decreasing in velocity.
Tuesday, July 20, 2010
3.3: Product and Quotient Rules
The product and quotient rules are both really easy. I actually really don't mind solving derivatives. They're not that bad..
Problem:
1. Find the derivative of ((10x^2 - e^x)(15x + 9^x)) / (5*4th root of x + x^7)
This problem uses a combination of rules. First apply the quotient rule:
(5*4th root of x + x^7)[(10x^2-e^x)(15x+9^x)]' - (10x^2-e^x)(15x+9^x)(5*4th root of x+x^7)' all divided by (5*4th root of x + x^7)^2.
Then I applied the product rule to find the derivative of [(10x^2-e^x)(15x+9^x)] which is (10x^2-e^x)(15x+9^x)' + (15x+9^x)(10x^2-e^x)'
I also used the power rule to find the derivative where exponents were present. All put together, the derivative of the above function is:
(5*4th root of x + x^7)(10x-e^x)(15+ln9*9^x)+(15x+9^x)(20x-e^x)-(10x^2-e^x)(15x+9^x)((5/4)x^(-3/4)+7x^6) ALL divided by (5*4th root of x + x^7)^2
Problem:
1. Find the derivative of ((10x^2 - e^x)(15x + 9^x)) / (5*4th root of x + x^7)
This problem uses a combination of rules. First apply the quotient rule:
(5*4th root of x + x^7)[(10x^2-e^x)(15x+9^x)]' - (10x^2-e^x)(15x+9^x)(5*4th root of x+x^7)' all divided by (5*4th root of x + x^7)^2.
Then I applied the product rule to find the derivative of [(10x^2-e^x)(15x+9^x)] which is (10x^2-e^x)(15x+9^x)' + (15x+9^x)(10x^2-e^x)'
I also used the power rule to find the derivative where exponents were present. All put together, the derivative of the above function is:
(5*4th root of x + x^7)(10x-e^x)(15+ln9*9^x)+(15x+9^x)(20x-e^x)-(10x^2-e^x)(15x+9^x)((5/4)x^(-3/4)+7x^6) ALL divided by (5*4th root of x + x^7)^2
3.2: The Derivative as a Function
Probably the most interesting part of this section is the way the power rule works to find the derivative. I feel like this section is pretty straightforward and really just builds on section 3.1 giving more tools to make solving derivatives even easier.
Problem:
1. Find the derivative of (7x^9 + square root x + 3x^(5/8)).
I used a combination of the power rule and sum rule to solve this derivative. The sum rule says that you can take the derivative of each piece of the function individually. Starting with 7x^9, I used the power rule to move the 9 down in front of the x, then I subtracted 1 from the exponent 9. This gave me 7*9x^8. The square root of x is equal to x^(1/2) so I used that instead, and applied the power rule. This gave me 1/2x^(-1/2). For 3x^(5/8) I used the power rule again, moving 5/8 down in front of x and subtracting 1 from 5/8. This gave me 3*(5/8)x^(-3/8). Put together, the derivative of this function is 63x^8 + (1/2)x^(-1/2) + (15/8)x^(-3/8)
Problem:
1. Find the derivative of (7x^9 + square root x + 3x^(5/8)).
I used a combination of the power rule and sum rule to solve this derivative. The sum rule says that you can take the derivative of each piece of the function individually. Starting with 7x^9, I used the power rule to move the 9 down in front of the x, then I subtracted 1 from the exponent 9. This gave me 7*9x^8. The square root of x is equal to x^(1/2) so I used that instead, and applied the power rule. This gave me 1/2x^(-1/2). For 3x^(5/8) I used the power rule again, moving 5/8 down in front of x and subtracting 1 from 5/8. This gave me 3*(5/8)x^(-3/8). Put together, the derivative of this function is 63x^8 + (1/2)x^(-1/2) + (15/8)x^(-3/8)
Monday, July 19, 2010
3.1: Definition of a Derivative
Finding the derivative is basically just finding the slope of a line tangent to a point on a curve. I think it is actually pretty easy to find the derivative of something. I think a lot of people hear the word "derivative" and talk themselves into thinking that its this really impossible math concept that they'll never understand and thats why its so hard for them to solve even simple problems. But I do think its confusing that there are so many different ways to write derivative and it can also be confusing to know which equation will work best when trying to solve a derivative problem.
Problem:
1. Find the derivative of f(x)= 3x^2 - 2x + 5
f'(x) = lim as h approaches 0 of f(x+h) - f(x)
-------------
h
Plugging in the numbers for x and h gave me: f'(x) = lim as h approaches 0 of
3(x+h)^2 - 2(x+h) + 5 - (2x^2 + 2x + 5)
------------------------------------------------
h
Next I added like terms and did simple algebra to get: lim as h approaches 0 of 6ah + 3h^2 -2h
-------------------
h
Then I factored out h so I had: lim as h approaches 0 of
h(6a + 3h -2)
-------------------
h
The h's cancel out, because they are present in both the numerator and denominator, so I was left with the limit as h approaches 0 of (6a + 3h -2).
Then I plugged in 0 for the last remaining h, which left me with the limit as h approaches 0 = 6a-2
Problem:
1. Find the derivative of f(x)= 3x^2 - 2x + 5
f'(x) = lim as h approaches 0 of f(x+h) - f(x)
-------------
h
Plugging in the numbers for x and h gave me: f'(x) = lim as h approaches 0 of
3(x+h)^2 - 2(x+h) + 5 - (2x^2 + 2x + 5)
------------------------------------------------
h
Next I added like terms and did simple algebra to get: lim as h approaches 0 of 6ah + 3h^2 -2h
-------------------
h
Then I factored out h so I had: lim as h approaches 0 of
h(6a + 3h -2)
-------------------
h
The h's cancel out, because they are present in both the numerator and denominator, so I was left with the limit as h approaches 0 of (6a + 3h -2).
Then I plugged in 0 for the last remaining h, which left me with the limit as h approaches 0 = 6a-2
Sunday, July 18, 2010
2.8: The Formal Definition of a Limit
The formal definition of a limit is probably one of the most confusing things I have ever read. I read the definition about 5 times and even after the discussion in class I still don't really understand what it means. I prefer my own definition.
Real life example of a limit: Finding the acceleration of a moving object.
This is an example of a limit because as the object moves it is getting closer and closer to its final position, the same way a limit gets closer and closer to a specific number.
Problems:
1. Prove that the lim of 5x+2 = 22 as x approaches 4 by relating the gap.
Set f(x)-22 = |(5x+2)-22| = |5x-20| = 5*|x-4|, this means that the gap is 5 times as large as |x-4|.
2. Find a value for δ so that ε=0.4 for the function lim 10x+15 = 105 as x approaches 9.
First solve the limit by relating the gap: f(x) - 105 = |(10x+15)-105| = |10x-90| = 10|x-90|. Then set up the equation so that|x-90| < ε/10. Plug in 0.4, so 0.4/10 = 0.04. Now the gap has to be less than 0.04. So, |f(x)-105|<0.4 if 0<|x-90|<0.04
Real life example of a limit: Finding the acceleration of a moving object.
This is an example of a limit because as the object moves it is getting closer and closer to its final position, the same way a limit gets closer and closer to a specific number.
Problems:
1. Prove that the lim of 5x+2 = 22 as x approaches 4 by relating the gap.
Set f(x)-22 = |(5x+2)-22| = |5x-20| = 5*|x-4|, this means that the gap is 5 times as large as |x-4|.
2. Find a value for δ so that ε=0.4 for the function lim 10x+15 = 105 as x approaches 9.
First solve the limit by relating the gap: f(x) - 105 = |(10x+15)-105| = |10x-90| = 10|x-90|. Then set up the equation so that|x-90| < ε/10. Plug in 0.4, so 0.4/10 = 0.04. Now the gap has to be less than 0.04. So, |f(x)-105|<0.4 if 0<|x-90|<0.04
2.7: Intermediate Value Theorem
The Intermediate Value Theorem is pretty straightforward. A continuous function can not skip values - meaning that if the function went from 10 to 200,000 it would hit every single value inbetween those numbers. This seems really obvious to me and I'm not sure why they would dedicate an entire section just to that. The Bisection Method, used to find zeros within specific intervals is also really straightforward.
Problems:
1. Show that the function f(x)= x^2 / x^7+8 takes on the value 0.10. To solve this problem I chose a pretty obvious interval: [0,1]. Plugging in 0 for x, gave me 0^2/ 0^7+8 which equals 0/8 or 0. So obviously there is a zero in this interval, but since the question isn't asking to find a zero, I plugged in 1 for x. This gave me 1^2/ 1^7 +8 = 1/9. 1/9 converted to a decimal = 0.1 repeating, which is really, really close to 0.10.
2. Carry out the Bisection Method for cos θ - 2*sinθ to determine whether or not there are zeros in the interval [1.25, 1.5].
There can't be a zero in this interval because if you plug 1.25 in for θ, you get -1.58. If you plug 1.5 in for θ, you get -1.92. From -1.58 to -1.92 there are no zeros. However, there are zeros in the interval [3.5,3.75]. When 3.5 is plugged in for θ the answer is -0.2349 and when 3.75 is plugged in for θ you get 0.3226. Since the answers go from negative to positive there has to be a zero in that interval.
Problems:
1. Show that the function f(x)= x^2 / x^7+8 takes on the value 0.10. To solve this problem I chose a pretty obvious interval: [0,1]. Plugging in 0 for x, gave me 0^2/ 0^7+8 which equals 0/8 or 0. So obviously there is a zero in this interval, but since the question isn't asking to find a zero, I plugged in 1 for x. This gave me 1^2/ 1^7 +8 = 1/9. 1/9 converted to a decimal = 0.1 repeating, which is really, really close to 0.10.
2. Carry out the Bisection Method for cos θ - 2*sinθ to determine whether or not there are zeros in the interval [1.25, 1.5].
There can't be a zero in this interval because if you plug 1.25 in for θ, you get -1.58. If you plug 1.5 in for θ, you get -1.92. From -1.58 to -1.92 there are no zeros. However, there are zeros in the interval [3.5,3.75]. When 3.5 is plugged in for θ the answer is -0.2349 and when 3.75 is plugged in for θ you get 0.3226. Since the answers go from negative to positive there has to be a zero in that interval.
Wednesday, July 14, 2010
2.6 Trigonometric Limits
The squeeze theorem can be used to find the limit of functions inbetween other functions, by squeezing the original function in. I think the way the squeeze theorem works is pretty cool.
Problems:
1. Use the squeeze theorem to find the lim f(x) as x approaches 3, when lim 3x+2 ≤ f(x) ≤ x^2 + 2.
To solve this problem I found the limit of the function on either side of f(x). The function 3x+2 is a continuous function, so the limit of 3x+2 when x approaches 3 can be found by plugging in 3 for x. When I plugged in 3, I got the limit of 3x+2 as x approaches 3 = 11. To find the limit of x^2 + 2 I did the same thing. X^2 + 2 is also a continuous function, so you can plug 3 in for x again. The limit of x^2 + 2 as x approaches 3 is also equal to 11. Therefore, the limit of f(x) as x approaches 3 is also 11.
2. Evaluate the limit: lim (1-cos(2θ)) / (sin 6θ) as θ approaches 0.
The expression 1-cosθ is equal to sinθ, so I changed 1-cosθ to sinθ, leaving me with sinθ over sinθ which is equal to 1. When θ approaches 0, (1-cosθ/θ) is equal to zero. If you multiply (2/6) * 1* 0, the answer is 0, therefore the limit of (1-cos(2θ)) / (sin6θ) as x approaches 0 = 0.
Problems:
1. Use the squeeze theorem to find the lim f(x) as x approaches 3, when lim 3x+2 ≤ f(x) ≤ x^2 + 2.
To solve this problem I found the limit of the function on either side of f(x). The function 3x+2 is a continuous function, so the limit of 3x+2 when x approaches 3 can be found by plugging in 3 for x. When I plugged in 3, I got the limit of 3x+2 as x approaches 3 = 11. To find the limit of x^2 + 2 I did the same thing. X^2 + 2 is also a continuous function, so you can plug 3 in for x again. The limit of x^2 + 2 as x approaches 3 is also equal to 11. Therefore, the limit of f(x) as x approaches 3 is also 11.
2. Evaluate the limit: lim (1-cos(2θ)) / (sin 6θ) as θ approaches 0.
The expression 1-cosθ is equal to sinθ, so I changed 1-cosθ to sinθ, leaving me with sinθ over sinθ which is equal to 1. When θ approaches 0, (1-cosθ/θ) is equal to zero. If you multiply (2/6) * 1* 0, the answer is 0, therefore the limit of (1-cos(2θ)) / (sin6θ) as x approaches 0 = 0.
2.5: Evaluating Limits Algebraically
Section 2.5 shows how to evaluate limits algebraically. I think that in some cases it is easier to solve algebraically, and it also saves some time rather than using the table method. However, I am not very good at math and usually make a lot of mistakes when trying to solve something algebraically so I will probably never use this method unless using a table is impossible.
Problems:
1. Solve this limit using algebra: lim (x^2 - 100) / (x-10) as x approaches 10.
I factored the numerator into (x+10)(x-10) / (x-10). Since (x-10) is present in both the numerator and the denominator, I cancelled it out, leaving me with the limit (x+10) as x approaches 10. Plugigng 10 in for x gave me the limit of (10+10) as x approaches 10 = 20.
2. Solve this limit: lim cot θ / csc θ as θ approaces 0.
Before I started to solve this problem, I changed cot and csc so that they were expressed in terms of sin and cos. Cot θ is equal to cos θ / sin θ and csc θ is equal to 1/sinθ. Since csc = 1/sinθ and is located on the denominator, I multiplied the entire limit by the reciprocal, sin θ / 1. That gave me: lim cosθ*sinθ / sinθ as x approaches 0. Because sinθ is present in both the numerator and denominator, you can cross it out. This leaves only lim cosθ as x approaches 0. If you plug in 0 for θ you get the lim cosθ as x approaches 0 = 1.
Problems:
1. Solve this limit using algebra: lim (x^2 - 100) / (x-10) as x approaches 10.
I factored the numerator into (x+10)(x-10) / (x-10). Since (x-10) is present in both the numerator and the denominator, I cancelled it out, leaving me with the limit (x+10) as x approaches 10. Plugigng 10 in for x gave me the limit of (10+10) as x approaches 10 = 20.
2. Solve this limit: lim cot θ / csc θ as θ approaces 0.
Before I started to solve this problem, I changed cot and csc so that they were expressed in terms of sin and cos. Cot θ is equal to cos θ / sin θ and csc θ is equal to 1/sinθ. Since csc = 1/sinθ and is located on the denominator, I multiplied the entire limit by the reciprocal, sin θ / 1. That gave me: lim cosθ*sinθ / sinθ as x approaches 0. Because sinθ is present in both the numerator and denominator, you can cross it out. This leaves only lim cosθ as x approaches 0. If you plug in 0 for θ you get the lim cosθ as x approaches 0 = 1.
2.4: Limits and Continuity
Continuity is fairly easy to understand. A function whose graph does not have any breaks in it will be continuous. Discontinuities are a little more complicated, but not by much. The only thing that makes discontinuities slightly more confusing is the different types: jump, removable and infinite.
Problems:
1. Using the substitution method, evaluate the limit of cos^(-1) (x/9) as x approaches 9.
To solve I plugged 9 in for x giving me the limit of cos^(-1) (9/9) as x approaches 9, which is also equal to lim cos^(-1) (1) as x approaches 9 = 0.
2. What is the domain of the function and is it continuous on its domain? f(x) = square root of (x^2+64)
For this problem x can be any number greater than or equal to 8. For any number less than 8, the function will be discontinuous.
Problems:
1. Using the substitution method, evaluate the limit of cos^(-1) (x/9) as x approaches 9.
To solve I plugged 9 in for x giving me the limit of cos^(-1) (9/9) as x approaches 9, which is also equal to lim cos^(-1) (1) as x approaches 9 = 0.
2. What is the domain of the function and is it continuous on its domain? f(x) = square root of (x^2+64)
For this problem x can be any number greater than or equal to 8. For any number less than 8, the function will be discontinuous.
2.3: Basic Limit Laws
Section 2.3 is a short section that just outlines the rules that must be followed for limits. I've never understood how to solve limits until now, so I never knew what the limit laws were or how to use them. The 'laws' basically just show you an easier approach to a problem depending on whether it is an addition/subtraction, or multiplication/division problem.
Problems:
1. Assume that lim f(x) as x approaches 4 = 2 and the lim g(x) as x approaches 4 = 5. Evaluate the limit: lim(3*f(x) = 3*g(x)) as x approaches 4.
To solve this problem, I broke it up into the limit of 3 * f(x) + the limit of 3 * g(x) both as x approaches 4. Plugging the given numbers in for f(x) and g(x), I had: the limit of 3*2 as x approaches 4 + the limit of 3*5 as x approaches 4. Using the Constant Multiple Law, which basically says that if you have the limit of two numbers multiplied together you can move the first number to the other side of the limit equation so that you have 3*limit as x approaches 4 of f(x). The Constant Multiple Law gave me: 3*limit of 2 as x approaches 4 =6 and 3* the limit of 5 as x approaches 4 = 15.
2. Is it possible to use the Quotient Law to evalue the limit of tan x/ x as x approaches 0?
Separate this problem using the Quotient Law, it becomes limit of tan x as x approaches 0 divided by the limit of x as x approaches 0. If you were to just plug 0 in for x, the equation would be undefined, meaning that you would have to find another way to solve it. If you were trying to solve this problem, you could use the graphical or numerical method to solve. I used the numerical method and found that the limit was 1, just to make sure that it was at all possible to solve.
Problems:
1. Assume that lim f(x) as x approaches 4 = 2 and the lim g(x) as x approaches 4 = 5. Evaluate the limit: lim(3*f(x) = 3*g(x)) as x approaches 4.
To solve this problem, I broke it up into the limit of 3 * f(x) + the limit of 3 * g(x) both as x approaches 4. Plugging the given numbers in for f(x) and g(x), I had: the limit of 3*2 as x approaches 4 + the limit of 3*5 as x approaches 4. Using the Constant Multiple Law, which basically says that if you have the limit of two numbers multiplied together you can move the first number to the other side of the limit equation so that you have 3*limit as x approaches 4 of f(x). The Constant Multiple Law gave me: 3*limit of 2 as x approaches 4 =6 and 3* the limit of 5 as x approaches 4 = 15.
2. Is it possible to use the Quotient Law to evalue the limit of tan x/ x as x approaches 0?
Separate this problem using the Quotient Law, it becomes limit of tan x as x approaches 0 divided by the limit of x as x approaches 0. If you were to just plug 0 in for x, the equation would be undefined, meaning that you would have to find another way to solve it. If you were trying to solve this problem, you could use the graphical or numerical method to solve. I used the numerical method and found that the limit was 1, just to make sure that it was at all possible to solve.
Tuesday, July 13, 2010
2.2: Limits: A Numerical and Graphical Approach
The whole concept of limits is something that I've never really understood. I first learned about limits in pre-calculus, but did not understand how they worked or even what they were. I did not know that there were different types of limits. I didn't realize that approaching a number from either the left or right side of that number could possibly yield a different limit. I also didn't know that the limit of a function only exists if both of the one-sided limits exist and are equal.
Real life example of a limit: compounding interest rates. This is an example of a limit because as the rate is continuously compounded it only adds a very small amount of money to your bank account. To actually make a significant amount of money from the interest alone, that money would have to sit untouched for an extremely long time because the amount compounded each time is so small.
Problems:
1. Decide whether the limit as x approaches 0 of cos3x / x exists or does not exist. If it exists, find the limit.
Using the table method for finding a limit I found that this limit does not exist. When approaching x from the right and from the left the limit does not exist.
2. What is the limit of 2x+4 as x approaches 2?
The limit is 8. This function is continuous, because is it is a linear function, therefore you can just plug 2 in for x to solve the equation.
Real life example of a limit: compounding interest rates. This is an example of a limit because as the rate is continuously compounded it only adds a very small amount of money to your bank account. To actually make a significant amount of money from the interest alone, that money would have to sit untouched for an extremely long time because the amount compounded each time is so small.
Problems:
1. Decide whether the limit as x approaches 0 of cos3x / x exists or does not exist. If it exists, find the limit.
Using the table method for finding a limit I found that this limit does not exist. When approaching x from the right and from the left the limit does not exist.
2. What is the limit of 2x+4 as x approaches 2?
The limit is 8. This function is continuous, because is it is a linear function, therefore you can just plug 2 in for x to solve the equation.
Monday, July 12, 2010
2.1: Limits, Rates of Change and Tangent Lines
This section mostly dealt with rates of change. While I have been familiar with rates of change for years as a result of many math classes, I didn't realize how useful they are in real life. The paragraph regarding instantaneous rates of change in velocity was really interesting, as was the example that used the average rate of change to determine the speed of sound in air. I think its really interesting how math and science, more specifically physics in this problem go together.
Problems:
1. Estimate the instantaneous velocity of a particle traveling a distance of s(t) = t^2 + 4t at t=5, using intervals to the left and right of t=5.
To solve this I used 4.99 to approach 5 from the left and 5.01 to approach 5 on the right. I set my equations up to look like this:
[4.99,5]: 5^2 + 2(5) - (4.99)^2 - 2(4.99) all divided by 0.01.
[5,5.01]: 5.01^2 + 2(5.01) - 5^2 - 2(5) all divided by 0.01.
The reason I divided each by 0.01 is because that is the difference between the intervals in both equations. The answers that I got were 11.99 for the first equation and 12.01 for the second equation. This means that when t=5, the instantaneous rate of change is approximately 12.
2. Estimate the average rate of change of cos x, at [π/6, π/4].
cos(π/4) - cos(π/6)
----------------- = -0.60701
(π/4-(π/6)
Problems:
1. Estimate the instantaneous velocity of a particle traveling a distance of s(t) = t^2 + 4t at t=5, using intervals to the left and right of t=5.
To solve this I used 4.99 to approach 5 from the left and 5.01 to approach 5 on the right. I set my equations up to look like this:
[4.99,5]: 5^2 + 2(5) - (4.99)^2 - 2(4.99) all divided by 0.01.
[5,5.01]: 5.01^2 + 2(5.01) - 5^2 - 2(5) all divided by 0.01.
The reason I divided each by 0.01 is because that is the difference between the intervals in both equations. The answers that I got were 11.99 for the first equation and 12.01 for the second equation. This means that when t=5, the instantaneous rate of change is approximately 12.
2. Estimate the average rate of change of cos x, at [π/6, π/4].
cos(π/4) - cos(π/6)
----------------- = -0.60701
(π/4-(π/6)
Sunday, July 11, 2010
1.6 Exponential and Logarithmic Functions
Exponential and logarithmic functions are inverses of each other. Exponential functions require bases which are raised to some power. These functions are always positive, but not equal to one to avoid producing a constant function. Exponential functions increase rapidly. This is proved by the most important exponent law, which states that when two exponential functions with the same base are multiplied together the exponents are added. The number ‘e’ is approximately equal to 2.7 and is defined as being a “natural” base. Logarithmic functions also are defined with a base, but instead of being raised to a certain power, the log of that base, b is taken. The opposite of a log base ‘e’ is the natural log, ln. Hyperbolic functions create hyperbolas when graphed. Hyperbolic functions are related to trigonometric functions in that they are parities of each other. Hyperbolic functions are written just like trig functions, but with an ‘h’ added to the end. For example: cosh x.
Problems:
1. Rewrite as a whole number without using a calculator.
a) 7^0 =1
b) 10^2(2^-2 + 5^-2)=29
c) (4^3)^5/(4^5)^3 =1
d) 27^(4/3)=81
e) 8^(-1/3)*8^(5/3)=16
f) 3*4^(1/4)-12*2^(-3/2)=0
To solve this problem, I used the Laws of Exponents.
x1 = x
x0 = 1
x-1 = 1/x
xmxn = xm+n
xm/xn = xm-n
(xm)n = xmn
(xy)n = xnyn
(x/y)n = xn/yn
x-n = 1/xn
#40. Prove the addition formula for cosh x.
Cosh(x+y) = cosh x*cosh y + sinh x* sinh y
I know that cosh x is equal to ½(e^x + e^-x) and sinh x is equal to ½(e^x -e^-x). Using these values for cosh and sinh I plugged them into the addition formula and solved.
Problems:
1. Rewrite as a whole number without using a calculator.
a) 7^0 =1
b) 10^2(2^-2 + 5^-2)=29
c) (4^3)^5/(4^5)^3 =1
d) 27^(4/3)=81
e) 8^(-1/3)*8^(5/3)=16
f) 3*4^(1/4)-12*2^(-3/2)=0
To solve this problem, I used the Laws of Exponents.
x1 = x
x0 = 1
x-1 = 1/x
xmxn = xm+n
xm/xn = xm-n
(xm)n = xmn
(xy)n = xnyn
(x/y)n = xn/yn
x-n = 1/xn
#40. Prove the addition formula for cosh x.
Cosh(x+y) = cosh x*cosh y + sinh x* sinh y
I know that cosh x is equal to ½(e^x + e^-x) and sinh x is equal to ½(e^x -e^-x). Using these values for cosh and sinh I plugged them into the addition formula and solved.
1.5: Inverse Functions
An inverse function is a function that has a reverse effect on the original function. For example, the inverse of addition is subtraction and vice versa. Invertible functions are functions that are one to one on their domains. This means that the domain of a function ‘f’ is equal to the range of its inverse, commonly known as ‘f-1.’ The range of ‘f’ is equal to the domain of the inverse function. The horizontal line test can be used to determine if a function is one to one. The horizontal line test is performed the same way as the vertical line test, except that a line drawn horizontally across the graph should only intersect one point. A function can be made into a one to one function by restricting the domain and using a reflection of the original line, which would be the same as the inverse. Trigonometric functions are not one to one, but can be made into one to one functions by restricting their domains. The inverse of trigonometric functions are named by adding “arc” to the front of the trig function, for example: “arcsin” or by simply saying sine inverse.
Problems:
#1. Show that f(x)= 7x-4 is invertible by finding its inverse.
The inverse of f(x) = 1/7(x-4). To solve this problem I made 7 into a fraction equal to 7/1 and then I flipped that function to give me 1/7. The rest of thequation, (x-4) only needed parentheses added to it to show that in order for the function to work correctly, the 4 must be subtracted from the value of x and that answer should be multiplied by 1/7.
#40. Simplify by referring to the appropriate triangle or trigonometric identity. Cos(tan-1 x).
To solve this problem I drew a right triangle on my paper. Using the Pythagorean Theorem I labeled the sides of the triangle with the far left side equal to√(1-x^2 ), the bottom of the triangle equal to x and the long right side equal to 1. If the angle on the bottom right side is set to be θ, which would make the trigonometric function ‘tan x’ true, the equation can be solved for cos θ. Cos θ is equal to the adjacent over the hypotenuse, in this case the answer would be x/1 or just x.
Problems:
#1. Show that f(x)= 7x-4 is invertible by finding its inverse.
The inverse of f(x) = 1/7(x-4). To solve this problem I made 7 into a fraction equal to 7/1 and then I flipped that function to give me 1/7. The rest of thequation, (x-4) only needed parentheses added to it to show that in order for the function to work correctly, the 4 must be subtracted from the value of x and that answer should be multiplied by 1/7.
#40. Simplify by referring to the appropriate triangle or trigonometric identity. Cos(tan-1 x).
To solve this problem I drew a right triangle on my paper. Using the Pythagorean Theorem I labeled the sides of the triangle with the far left side equal to√(1-x^2 ), the bottom of the triangle equal to x and the long right side equal to 1. If the angle on the bottom right side is set to be θ, which would make the trigonometric function ‘tan x’ true, the equation can be solved for cos θ. Cos θ is equal to the adjacent over the hypotenuse, in this case the answer would be x/1 or just x.
1.4:Trigonometric Functions
This section describes the relationship between radians and degrees. To convert from radians to degrees multiply the number by 180/pi, to convert the other way, from degrees to radians multiply by the reciprocal, pi/180. The section also explains how trigonometric functions are defined using right triangles. The Pythagorean Theorem is important to finding the values of the trigonometric functions in right triangles as well as in identities.
Problems:
#4. Convert from degrees to radians.
a) 1° =0.0175 radians
b) 30°=0.5236 radians
c) 25°=0.4363 radians
d) 120°=2.094 radians
To solve this problem I used my calculator and the formula for converting from degrees to radians. I multiplied each number in degrees by pi/180 to give the number of radians.
#38. Solve sinθ=cos 2θ for 0≤θ<2pi.
To solve this problem I substituted cos2θ for cos^2θ-sin^2θ, giving me sinθ= cos^2θ-sin^2θ. Then I set the whole equation to zero; sinθ -cos^2θ-sin^2θ=0. Using the double angle formulas, I changed the equation to sinθ-((1/2)*1 + cos2x) –((1/2)*1 – cos2x) = 0. Solving for sinθ, I found it to be equal to ½ and -1. On a unit circle, ½ corresponds to pi/6 and 5pi/6 and -1 corresponds to 3pi/2.
Problems:
#4. Convert from degrees to radians.
a) 1° =0.0175 radians
b) 30°=0.5236 radians
c) 25°=0.4363 radians
d) 120°=2.094 radians
To solve this problem I used my calculator and the formula for converting from degrees to radians. I multiplied each number in degrees by pi/180 to give the number of radians.
#38. Solve sinθ=cos 2θ for 0≤θ<2pi.
To solve this problem I substituted cos2θ for cos^2θ-sin^2θ, giving me sinθ= cos^2θ-sin^2θ. Then I set the whole equation to zero; sinθ -cos^2θ-sin^2θ=0. Using the double angle formulas, I changed the equation to sinθ-((1/2)*1 + cos2x) –((1/2)*1 – cos2x) = 0. Solving for sinθ, I found it to be equal to ½ and -1. On a unit circle, ½ corresponds to pi/6 and 5pi/6 and -1 corresponds to 3pi/2.
1.3: The Basic Classes of Functions
This section outlines the different types of important functions in calculus. There are polynomials, which have whole number exponents and consist of a bunch of numbers added together. There are also rational functions, defined by P(x)/Q(x) where Q(x) is not equal to zero, because it would make the function undefined. Algebraic functions involve a variety of operations, polynomials and rational functions. They are kind of like a combination of the other types of functions outlined in this section. Exponential functions are functions that are always positive and involve bases raised to a specific exponent. They are the inverse of Logs, or logarithmic functions, which can be used to find an unknown exponent. Both exponential and logarithmic functions are found in everyday life, some examples being the Richter scale, pH levels, the economy, C13 dating, etc. Section 1.3 also describes how to construct new functions. New functions can be formed by adding functions, subtracting them, multiplying and dividing them. Composition of functions occurs when you have a value for f(x) and a value for g(x) and you combine them by plugging the value of g(x) in for the value of x in f(x). For example if f(x) = x^2 and g(x) = 5-x and you want to find out what f(g(x)) is equal to; you would write (5-x)^2 and solve for x. This can also be turned around to find g(f(x)).
Square root of 1-t ^3
Problems:
#34: Calculate the composite functions f(g(x)) and g(f(x)) and determine their domains.
F(t) = √t g(x) = 1-t^3
F(g(x)= 〖√(1-t)〗^3D: DNE for any value of t. g(f(x))=1-√t ^3 D: (0, infinity)
To solve this problem, I started with f(g(x)). I plugged the value of g(x) in for t in the function f(x). To solve g(f(x)) I did the same thing, but opposite, I plugged the value of f(x) in for t in the function g(x). To determine the domains of the functions I tried various numbers for t and carried out the function. I found that every value I tried for f(g(x)) left me with a non real answer, allowing me to come to the conclusion that there was no possible value for t which would make this function work. For the function g(f(x)) I did the same thing, using random values for t. I found that negative numbers would not work, because you cannot take the square root of a negative number, so I concluded that the domain would have to be from zero to infinity.
#36: Find all values of c such that the domain of f(x)= (x+1)/(x^2+2cx+4) is all real numbers.
I think that all real numbers will work when plugged in for c to keep the domain of the function all real numbers. I plugged a variety of numbers in for c, using x =2 and every value that I found fit into the category all real numbers.
Square root of 1-t ^3
Problems:
#34: Calculate the composite functions f(g(x)) and g(f(x)) and determine their domains.
F(t) = √t g(x) = 1-t^3
F(g(x)= 〖√(1-t)〗^3D: DNE for any value of t. g(f(x))=1-√t ^3 D: (0, infinity)
To solve this problem, I started with f(g(x)). I plugged the value of g(x) in for t in the function f(x). To solve g(f(x)) I did the same thing, but opposite, I plugged the value of f(x) in for t in the function g(x). To determine the domains of the functions I tried various numbers for t and carried out the function. I found that every value I tried for f(g(x)) left me with a non real answer, allowing me to come to the conclusion that there was no possible value for t which would make this function work. For the function g(f(x)) I did the same thing, using random values for t. I found that negative numbers would not work, because you cannot take the square root of a negative number, so I concluded that the domain would have to be from zero to infinity.
#36: Find all values of c such that the domain of f(x)= (x+1)/(x^2+2cx+4) is all real numbers.
I think that all real numbers will work when plugged in for c to keep the domain of the function all real numbers. I plugged a variety of numbers in for c, using x =2 and every value that I found fit into the category all real numbers.
1.2: Linear and Quadratic Functions
Linear functions are those that form a line when graphed. A linear function is one of the most widely known types of functions, with the formula y=mx+b. The equation for a linear function can be found using point slope form: y-b=m(x-a) and point-point form: y-b1=m(x-a1). The slope of the line, also known as the change in y over the change in x or the rise over the run is indicated by the variable ‘m.’ A function known as a quadratic function is a polynomial. In a quadratic function f(x)=ax^2+bx+c and the graph represents a parabola, which opens either upward or downward depending on the sign of the coefficient, a. To find the roots of a function you can use either the quadratic formula or you can complete the square.
Problems:
#33: Find the roots of the quadratic polynomials.
a) 4x^2-3x-1
To find the roots of this equation I plugged the numbers into the quadratic formula, The answers that I came up with were +1 and -0.25.
b) X^2-3x-1
I solved this problem exactly the same way as the above problem. My answers were 2.41 and -0.414
#53: Show that if f(x) and g(x) are linear, then so is f(x) + g(x). Is the same true of f(x)g(x)?
The function f(x) +g(x) will still be linear if both f(x) and g(x) are linear. This is true because the addition of the two functions will still create lines, although they may be either perpendicular or parallel depending on the exact equation that defines them. F(x)g(x) will also be linear if f(x) and g(x) are independently linear. To check my assumptions, I plugged random numbers in for f(x) and g(x) and graphed them using my calculator. I also graphed f(x) + g(x) and f(x)g(x) using those numbers. The graphs for all values were horizontal lines.
Problems:
#33: Find the roots of the quadratic polynomials.
a) 4x^2-3x-1
To find the roots of this equation I plugged the numbers into the quadratic formula, The answers that I came up with were +1 and -0.25.
b) X^2-3x-1
I solved this problem exactly the same way as the above problem. My answers were 2.41 and -0.414
#53: Show that if f(x) and g(x) are linear, then so is f(x) + g(x). Is the same true of f(x)g(x)?
The function f(x) +g(x) will still be linear if both f(x) and g(x) are linear. This is true because the addition of the two functions will still create lines, although they may be either perpendicular or parallel depending on the exact equation that defines them. F(x)g(x) will also be linear if f(x) and g(x) are independently linear. To check my assumptions, I plugged random numbers in for f(x) and g(x) and graphed them using my calculator. I also graphed f(x) + g(x) and f(x)g(x) using those numbers. The graphs for all values were horizontal lines.
1.1: Real Numbers, Functions and Graphs
This section describes real numbers as being either finite, infinite but not repeating and repeating. A finite number is a number such as 1/2 which when converted to decimal form is only 0.5. An infinite but not repeating number is a number such as pi, where the decimal places continue on to infinity, but the sequence is never repeated and a repeating number is one where a certain sequence of numbers is repeated over and over again. Numbers can also be defined as rational and irrational. A rational number will be finite or repeating, while an irrational number will be infinite but not repeating. Intervals are used to understand how numbers relate to one another. There are closed intervals, denoted by brackets [ ] which include the end numbers; open intervals, denoted by parentheses ( ) which exclude the end numbers and half open intervals, denoted by a bracket and a parenthesis [ ) or ( ] which either include or exclude the number on the end with the corresponding bracket or parenthesis.
Graphing is extremely helpful in calculus to help define the domain and range of equations. Producing a graph for a specific equation is helpful because it gives the person trying to solve the equation a clear image of what they are dealing with. A function is an equation that consists of inputs and outputs. Each input within a function can only have one specific output, making the individual inputs and outputs unique. Graphs of functions can be both shifted and scaled either vertically or horizontally depending on the way the equation is written.
Problems:
#29: Suppose that
x-4
is less than or equal to 1
a. What is the maximum possible value of
x+4
?
To find the answer to this problem I began to plug numbers in for ‘x.’ I used numbers starting with 1 and plugged them into the original equation to ensure that the answer would be less than or equal to 1. The largest possible number that I found which would work in the original equation was 5, with 5-4 = 1. This means that when plugged into the equation
x+4
, the maximum value for that equation would be 9.
b. Show that
x^2-16
is less than or equal to 9.
To solve this question I used the complete the square method. I broke up the equation into (x+4) and (x-4). I plugged in 5 for x, which is the largest possible value for x that would make this equation true. Using 5 for x, x^2 becomes 25 – 16 = 9. Any number plugged in for x that is less than or equal to 5 will make this equation true.
#48: Find the domain and range of the function g(t) = cos 1/t
To solve this problem, I used what I know about functions. I broke up the function into two parts: 1/t and cos x to help me figure the problem out. T could be any number other than zero because if t were equal to zero it would make the function undefined. The domain of the cosine function is from negative infinity to infinity which means that the domain for the entire function g(t)= cos 1/t should be from negative infinity to infinity. The range of the function 1/t is from (negative infinity, 0] and [0 to infinity). Zero is excluded because it would make the function undefined. The range of a cosine function is from (-1,1). When the two pieces of the function are combined I think that the range of the function g(t)= cos 1/t is from (-1, 0] and [0 to 1).
Graphing is extremely helpful in calculus to help define the domain and range of equations. Producing a graph for a specific equation is helpful because it gives the person trying to solve the equation a clear image of what they are dealing with. A function is an equation that consists of inputs and outputs. Each input within a function can only have one specific output, making the individual inputs and outputs unique. Graphs of functions can be both shifted and scaled either vertically or horizontally depending on the way the equation is written.
Problems:
#29: Suppose that
x-4
is less than or equal to 1
a. What is the maximum possible value of
x+4
?
To find the answer to this problem I began to plug numbers in for ‘x.’ I used numbers starting with 1 and plugged them into the original equation to ensure that the answer would be less than or equal to 1. The largest possible number that I found which would work in the original equation was 5, with 5-4 = 1. This means that when plugged into the equation
x+4
, the maximum value for that equation would be 9.
b. Show that
x^2-16
is less than or equal to 9.
To solve this question I used the complete the square method. I broke up the equation into (x+4) and (x-4). I plugged in 5 for x, which is the largest possible value for x that would make this equation true. Using 5 for x, x^2 becomes 25 – 16 = 9. Any number plugged in for x that is less than or equal to 5 will make this equation true.
#48: Find the domain and range of the function g(t) = cos 1/t
To solve this problem, I used what I know about functions. I broke up the function into two parts: 1/t and cos x to help me figure the problem out. T could be any number other than zero because if t were equal to zero it would make the function undefined. The domain of the cosine function is from negative infinity to infinity which means that the domain for the entire function g(t)= cos 1/t should be from negative infinity to infinity. The range of the function 1/t is from (negative infinity, 0] and [0 to infinity). Zero is excluded because it would make the function undefined. The range of a cosine function is from (-1,1). When the two pieces of the function are combined I think that the range of the function g(t)= cos 1/t is from (-1, 0] and [0 to 1).
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