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 +.