Chapter 2.1, Problem 74AYU

Given $f\left(x\right)\text{ = }\frac{1}{x}$ and $\left(\frac{f}{g}\right)\left(x\right)\text{ = }\frac{x\text{ + 1}}{x^2\text{ - }x}$, find the function $g$.

To determine

To find: The function $g$ if $f\left(x\right)=\frac{1}{x}$ and $\left(\frac{f}{g}\right)\left(x\right)=\frac{x+1}{x^2-x}$

Answer to Problem 74AYU

$g\left(x\right)=\frac{x-1}{x+1}$

Explanation of Solution

Given:

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

Calculation:

$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

$\frac{x+1}{x^2-x}=\frac{\frac{1}{x}}{g\left(x\right)}$

$g\left(x\right)=\left(\frac{1}{x}\right)\left(\frac{x^2-x}{x+1}\right)$

$g\left(x\right)=\frac{1}{x}\times \frac{x\left(x-1\right)}{x+1}$

$g\left(x\right)=\frac{x-1}{x+1}$