Chapter 1.5, Problem 43E

To determine

To find: The inverse of the given function and verify that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$.

Answer to Problem 43E

The $f^{-1}\left(x\right)={\log }_2\left(\frac{x}{100-x}\right)$ and it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$.

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{100}{1+2^{-x}}$.

Calculation:

Substitute $y$ for $f\left(x\right)$ in the equation $f\left(x\right)=\frac{100}{1+2^{-x}}$ and solve the equation for $x$ in terms of $y$.

  $\begin{array}{c}y=\frac{100}{1+2^{-x}}\\ 1+2^{-x}=\frac{100}{y}\\ 2^{-x}=\frac{100}{y}-1\end{array}$

Take log on both sides of the above equation.

  $\begin{array}{c}{\log }_2\left(2^{-x}\right)={\log }_2\left(\frac{100}{y}-1\right)\\ -x={\log }_2\left(\frac{100}{y}-1\right)\\ x=-{\log }_2\left(\frac{100-y}{y}\right)\\ ={\log }_2\left(\frac{y}{100-y}\right)\end{array}$

Interchange $x$ and $y$ , and substitute $y=f^{-1}\left(x\right)$.

  $\begin{array}{c}y={\log }_2\left(\frac{x}{100-x}\right)\\ f^{-1}\left(x\right)={\log }_2\left(\frac{x}{100-x}\right)\end{array}$

Verify:

Substitute $f^{-1}\left(x\right)$ for $x$ in the function $f\left(x\right)=\frac{100}{1+2^{-x}}$ to find $\left(f\circ f^{-1}\right)\left(x\right)$.

  $\begin{array}{c}\left(f\circ f^{-1}\right)\left(x\right)=f\left({\log }_2\frac{x}{100-x}\right)\\ =\frac{100}{1+2^{-{\log }_2\left(\frac{x}{100-x}\right)}}\\ =\frac{100}{1+2^{{\log }_2\left(\frac{100-x}{x}\right)}}\\ =\frac{100}{1+\frac{100-x}{x}}\end{array}$

Simplify further.

  $\begin{array}{c}\left(f\circ f^{-1}\right)\left(x\right)=\frac{100x}{x+100-x}\\ =\frac{100x}{100}\\ =x\end{array}$

Substitute $f\left(x\right)$ for $x$ in the function $f^{-1}\left(x\right)={\log }_2\left(\frac{x}{100-x}\right)$ to find $\left(f^{-1}\circ f\right)\left(x\right)$.

  $\begin{array}{c}\left(f^{-1}\circ f\right)\left(x\right)=f^{-1}\left(f\left(x\right)\right)\\ =f^{-1}\left(\frac{100}{1+2^{-x}}\right)\\ ={\log }_2\left(\frac{\frac{100}{1+2^{-x}}}{100-\frac{100}{1+2^{-x}}}\right)\\ ={\log }_2\left(\frac{100}{100\left(1+2^{-x}\right)-100}\right)\end{array}$

Simplify further.

  $\begin{array}{c}\left(f^{-1}\circ f\right)\left(x\right)={\log }_2\left(\frac{1}{2^{-x}}\right)\\ ={\log }_2\left(2^x\right)\\ =x\end{array}$

Therefore, the $f^{-1}\left(x\right)={\log }_2\left(\frac{x}{100-x}\right)$ and it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$.