Chapter 3, Problem 34RE

To determine

To calculate: The find all the values of x for which the function is differentiable.

Answer to Problem 34RE

The function is differentiable for all values of x except for $x=\frac{7}{2}$ .

Explanation of Solution

Given Information: The function is $y={\left(2x-7\right)}^{-1}\left(x+5\right)$ .

Calculation:

The given function is,

  $y={\left(2x-7\right)}^{-1}\left(x+5\right)$

The derivative of the function y with respect to x is calculated as,

  $\begin{array}{c}y={\left(2x-7\right)}^{-1}\left(x+5\right)\\ \frac{dy}{dx}=\frac{d}{dx}\left({\left(2x-7\right)}^{-1}\left(x+5\right)\right)\left(\text{Substitute the value }y\text{ in }\frac{dy}{dx}\right)\\ =\frac{d}{dx}\left(\frac{x+5}{2x-7}\right)\left(\text{Simplify the expression}\right)\\ =\frac{\left(2x-7\right)\frac{d}{dx}\left(x+5\right)-\left(x+5\right)\frac{d}{dx}\left(2x-7\right)}{{\left(2x-7\right)}^2}\left(\text{Apply quotient rule}\right)\\ =\frac{\left(2x-7\right)\times 1-\left(x+5\right)\times 2}{{\left(2x-7\right)}^2}\left(\because \frac{d}{dx}\left(x+5\right)=1,\frac{d}{dx}\left(2x-7\right)=2\right)\\ =\frac{2x-7-\left(2x+10\right)}{{\left(2x-7\right)}^2}\left(\text{Simplify}\right)\\ =\frac{2x-7-2x-10}{{\left(2x-7\right)}^2}\\ =\frac{-17}{{\left(2x-7\right)}^2}\end{array}$

The derivative of the function is not defined for that value of x where denominator is equal to 0. That is,

  $\begin{array}{c}2x-7=0\\ x=\frac{7}{2}\end{array}$

Hence, the function is differentiable for all values of x except for $x=\frac{7}{2}$ .

Conclusion:

The function is differentiable for all values of x except for $x=\frac{7}{2}$ .