Chapter 0.5, Problem 44E

To determine

The formula for $f^{-1}$ and the verification of $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Answer to Problem 44E

It has been determined that $f^{-1}\left(x\right)=\frac{\ln \left(\frac{x}{50-x}\right)}{\ln \left(1.1\right)}$ and it has been 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:

The function, $f\left(x\right)=\frac{50}{1+{1.1}^{-x}}$ .

Concept used:

The inverse of a function $f\left(x\right)$ can be obtained by assuming $y=f\left(x\right)$ and then solving for $x$ in terms of $y$ .

Calculation:

The given function is,

  $f\left(x\right)=\frac{50}{1+{1.1}^{-x}}$

Let $y=f\left(x\right)$ . Put this value in the above expression to get,

  $y=\frac{50}{1+{1.1}^{-x}}$

Simplifying,

  $\begin{array}{l}1+{1.1}^{-x}=\frac{50}{y}\\ {1.1}^{-x}=\frac{50}{y}-1\\ {1.1}^{-x}=\frac{50-y}{y}\\ {1.1}^x=\frac{y}{50-y}\end{array}$

On further simplification,

  $\begin{array}{l}\ln \left({1.1}^x\right)=\ln \left(\frac{y}{50-y}\right)\\ x\ln \left(1.1\right)=\ln \left(\frac{y}{50-y}\right)\\ x=\frac{\ln \left(\frac{y}{50-y}\right)}{\ln \left(1.1\right)}\end{array}$

As assumed previously,

  $y=f\left(x\right)$

Then,

  $x=f^{-1}\left(y\right)$

Put this value in $x=\frac{\ln \left(\frac{y}{50-y}\right)}{\ln \left(1.1\right)}$ to get,

  $f^{-1}\left(y\right)=\frac{\ln \left(\frac{y}{50-y}\right)}{\ln \left(1.1\right)}$

Replacing the variable $y$ by $x$ ,

  $f^{-1}\left(x\right)=\frac{\ln \left(\frac{x}{50-x}\right)}{\ln \left(1.1\right)}$

This is the required inverse function.

Now, as given and determined; $f\left(x\right)=\frac{50}{1+{1.1}^{-x}}$ and $f^{-1}\left(x\right)=\frac{\ln \left(\frac{x}{50-x}\right)}{\ln \left(1.1\right)}$ .

Then,

  $\begin{array}{c}\left(f\circ f^{-1}\right)\left(x\right)=f\left(f^{-1}\left(x\right)\right)\\ =f\left(\frac{\ln \left(\frac{x}{50-x}\right)}{\ln \left(1.1\right)}\right)\\ =\frac{50}{1+{1.1}^{-\frac{\ln \left(\frac{x}{50-x}\right)}{\ln \left(1.1\right)}}}\end{array}$

Simplifying,

  $\begin{array}{c}\left(f\circ f^{-1}\right)\left(x\right)=\frac{50}{1+{1.1}^{-\frac{\ln \left(\frac{x}{50-x}\right)}{\ln \left(1.1\right)}}}\\ =\frac{50}{1+{1.1}^{-{\log }_{1.1}\left(\frac{x}{50-x}\right)}}\\ =\frac{50}{1+{1.1}^{{\log }_{1.1}\left(\frac{50-x}{x}\right)}}\\ =\frac{50}{1+\left(\frac{50-x}{x}\right)}\end{array}$

On further simplification,

  $\begin{array}{c}\left(f\circ f^{-1}\right)\left(x\right)=\frac{50}{\frac{x+50-x}{x}}\\ =\frac{50x}{50}\\ =x\end{array}$

So, $\left(f\circ f^{-1}\right)\left(x\right)=x$ .

Similarly,

  $\begin{array}{c}\left(f^{-1}\circ f\right)\left(x\right)=f^{-1}\left(f\left(x\right)\right)\\ =f^{-1}\left(\frac{50}{1+{1.1}^{-x}}\right)\\ =\frac{\ln \left(\frac{\frac{50}{1+{1.1}^{-x}}}{50-\frac{50}{1+{1.1}^{-x}}}\right)}{\ln \left(1.1\right)}\end{array}$

Simplifying,

  $\begin{array}{c}\left(f^{-1}\circ f\right)\left(x\right)=\frac{\ln \left(\frac{\frac{1}{1+{1.1}^{-x}}}{1-\frac{1}{1+{1.1}^{-x}}}\right)}{\ln \left(1.1\right)}\\ =\frac{\ln \left(\frac{\frac{1}{1+{1.1}^{-x}}}{\frac{1+{1.1}^{-x}-1}{1+{1.1}^{-x}}}\right)}{\ln \left(1.1\right)}\\ =\frac{\ln \left(\frac{1}{{1.1}^{-x}}\right)}{\ln \left(1.1\right)}\\ =\frac{\ln \left({1.1}^x\right)}{\ln \left(1.1\right)}\end{array}$

On further simplification,

  $\begin{array}{c}\left(f^{-1}\circ f\right)\left(x\right)=\frac{x\ln \left(1.1\right)}{\ln \left(1.1\right)}\\ =x\end{array}$

So, $\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Thus, it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Conclusion:

It has been determined that $f^{-1}\left(x\right)=\frac{\ln \left(\frac{x}{50-x}\right)}{\ln \left(1.1\right)}$ and it has been verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .