Chapter 4.3, Problem 85E

a.

To determine

To define: to calculate the range of the function: $f\left(x\right)=\frac{2^x}{1+2^x}$

Answer to Problem 85E

Inverse function: $f^{-1}\left(x\right)={\log }_2\left(\frac{x}{1-x}\right)$ .

Explanation of Solution

Given information:

To calculate the range of the function: $f\left(x\right)=\frac{2^x}{1+2^x}$ .

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

  $y=\frac{2^x}{2^x+1}$

Multiply by $2^x+1$

  $\begin{array}{l}\left(2^x+1\right)y=2^x\\ y\cdot 2^x+y=2^x\\ y\cdot 2^x-2^x=-y\\ 2^x=\frac{y}{y-1}\end{array}$

Now further simplification of the function by the definition oflogarithm.

  $x={\log }_2\left(-\frac{y}{y-1}\right)={\log }_2\left(\frac{y}{1-y}\right)$

So here the inverse of the function is: $f^{-1}\left(x\right)={\log }_2\left(\frac{x}{1-x}\right)$

b.

To determine

To define:to calculate the range of the function: $f\left(x\right)=\frac{2^x}{1+2^x}$

Answer to Problem 85E

Inverse function : $f^{-1}\left(x\right)={\log }_2\left(\frac{x}{1-x}\right)$ .

Explanation of Solution

The function: $f\left(x\right)=\frac{2^x}{1+2^x}$

Given information:

To calculate the range of the function: $f\left(x\right)=\frac{2^x}{1+2^x}$

The definition of the logarithm gives :

  $\begin{array}{c}{\log }_{y}x=z\\ x=y^2\succ 0\end{array}$

For $y\succ 0,y\ne 1$

In this case : $f^{-1}\left(x\right)={\log }_2\left(\frac{x}{1-x}\right)$ .

So it must be : $\frac{x}{1-x}\succ 0$

So it must be $0\prec x\prec 1$ , and then $f^{-1}=\left(0,1\right)$ .