Chapter 5.2, Problem 8QR

To determine

To find : Where $f\left(x\right)=\frac{x}{x^2-1}$ is differentiable.

Answer to Problem 8QR

The function is differentiable for the interval $\left(-\infty ,-1\right)\cup \left(-1,\text{ 1}\right)\cup \left(1,\infty \right)$ .

Explanation of Solution

Given information :

The function is continuous for the interval $\left(-\infty ,-1\right)\cup \left(-1,\text{ 1}\right)\cup \left(1,\infty \right)$ .

Calculation :

Consider the given equation:

  $f\left(x\right)=\frac{x}{x^2-1}$

Since,

Hence, $f\left(x\right)=\frac{x}{x^2-1}$ if,

  $x^2-1\ne 0\text{ (Dinominator will not be zero)}$

Therefore,

  $\begin{array}{l}{x}^2-1\ne 0\\ \left(x-1\right)\left(x+1\right)\ne 0\\ \left(x-1\right)\left(x+1\right)\ne 0\\ x\ne 1,-1\end{array}$

So, the domain of the function is all real number except $1\text{ and}-1$ .

Since, the domain of the function is all real number except $1\text{ and}-1$ . Therefore, the function is continuous for all $x$ except $1\text{ and}-1$ .

Also, the function can only be differentiable at values where it is continuous. But, there may be value of where $f\left(x\right)$ is continuous but not differentiable.

Now check where $f\text{'}\left(x\right)$ is continuous. So, differentiate $f\left(x\right)$ with respect to $x$ .

  $\begin{array}{l}f\text{'}\left(x\right)=\frac{\left(x^2-1\right)1-x\left(2x\right)}{{\left(x^2-1\right)}^2}\\ f\text{'}\left(x\right)=\frac{x^2-1-2x^2}{{\left(x^2-1\right)}^2}\\ f\text{'}\left(x\right)=\frac{-x^2-1}{{\left(x^2-1\right)}^2}\end{array}$

Therefore, $f\text{'}\left(x\right)$ is also discontinuous at $x=\pm 1$ . So, $f\left(x\right)$ is differentiable for all real values of $x$ except $1\text{ and}-1$ .

Hence,

The function is differentiable for the interval $\left(-\infty ,-1\right)\cup \left(-1,\text{ 1}\right)\cup \left(1,\infty \right)$ .