Question

For the function = (7 – 8x)2, find ƒ–1. determine whether ƒ–1 is a function.

Answer 1

f-1 is (7-8x)2-1 yes it is the function

Answer 2

Answer:

The inverse of the function is $f^{-1}(x)=\frac{7\pm\sqrt{x}}{8}$ and it is not a function.

Step-by-step explanation:

The given function is

$f(x)=(7-8x)^2$

It can be written as

$y=(7-8x)^2$

Interchange x and y.

$x=(7-8y)^2$

$\pm\sqrt{x}=7-8y$

$8y=7-\pm\sqrt{x}$

Divide both sides by 8.

$y=\frac{7-\pm\sqrt{x}}{8}$

Put $y=f^{-1}(x)$

$f^{-1}(x)=\frac{7\pm\sqrt{x}}{8}$

It is not a function because for each value of x there exist more than one value of y.

Therefore the inverse of the function is $f^{-1}(x)=\frac{7\pm\sqrt{x}}{8}$ and it is not a function.