Chapter 2.3, Problem 6QR

To determine

To find:The functions $f$ and $f\circ g$ for $g\left(x\right)=\sqrt{x-1}$ and $\left(g\circ f\right)\left(x\right)=\frac{1}{x}$ , $x>0$ .

Answer to Problem 6QR

The function $f$ is $\frac{1}{x^2}+1$ and $f\circ g$ is $\frac{x}{x-1}$ for the given functions.

Explanation of Solution

Given information:The functions are $g\left(x\right)=\sqrt{x-1}$ and $\left(g\circ f\right)\left(x\right)=\frac{1}{x}$ .

Calculation:

Simplify the composite function $g\circ f$ .

  $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$

Substitute $f\left(x\right)$ for $x$ in the function $g\left(x\right)=\sqrt{x-1}$ .

  $\left(g\circ f\right)\left(x\right)=\sqrt{f\left(x\right)-1}$

As given the function $\left(g\circ f\right)\left(x\right)=\frac{1}{x}$ .

Substitute $\frac{1}{x}$ for $\left(g\circ f\right)\left(x\right)$ in the above equation.

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

So, the value of function $f$ is $\frac{1}{x^2}+1$ .

Simplify the composite function $f\circ g$ .

  $\begin{array}{c}\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(\sqrt{x-1}\right)\end{array}$

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

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

So, the value of $f\circ g$ is $\frac{x}{x-1}$ .

Therefore, the function $f$ is $\frac{1}{x^2}+1$ and $f\circ g$ is $\frac{x}{x-1}$ for the given functions.