Question

I need to verify identity functions for a one to one function.

Answer 1

We have the function

$f(x)=(x+6)^3$

1. For f^-1:

Let y = f(x) = (x+6)^3

Switch x and y to get:

$x=(y+6)^3$

And solve for y

$\begin{gathered} x^{\frac{1}{3}}=y+6 \\ x^{\frac{1}{3}}-6=y+6-6 \\ x^{\frac{1}{3}}-6=y \end{gathered}$

And we have y = f^-1(x)

Answer blank 1:

$f^{-1}(x)=x^{\frac{1}{3}}-6$

2. For f o f^-1 (x):

$(f\circ f^{-1})(x)=f(f^{-1}(x))$

And solve

$\begin{gathered} =f(x^{\frac{1}{3}}-6) \\ =(x^{\frac{1}{3}}-6+6)^3 \\ =(x^{\frac{1}{3}})^3 \\ =x \end{gathered}$

answer blank 2

$x^{\frac{1}{3}}-6$

answer blank 3

$x^{\frac{1}{3}}-6$

answer blank 4

$x^{\frac{1}{3}}$

3. For f^-1 o f:

$(f^{-1}\circ f)(x)=f^{-1}(f(x))$

Solve

$\begin{gathered} =f^{-1}((x+6)^3) \\ =\sqrt[3]{(x+6)^3}-6 \\ =x+6-6 \\ =x \end{gathered}$

answer blank 5

$(x+6)^3$

answer blank 6

$(x+6)^3$

answer blank 7

$x+6$