Sunday, July 11, 2010

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.

1 comment:

  1. Both of these solutions need work, and I will up the grade if you improve both. Ask me after a class for more if you do not see the errors.

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