Chapter 5.3, Problem 5QR

To determine

To find : The domain of $f$ and $f\text{'}$ .

Answer to Problem 5QR

The domain of $f$ and $f\text{'}$ is $\left(-\infty ,\infty \right)-\left\{2\right\}$ .

Explanation of Solution

Given information :

The given function is $f\left(x\right)=\frac{x}{x-2}$ .

Calculation:

The domain of the function is the values for which the function is defined.

The function $f\left(x\right)$ is not defined for

  $\begin{array}{l}x-2=0\\ x=2\end{array}$

Therefore, the domain of $f\left(x\right)=x^{3/5}$ is $\left(-\infty ,\infty \right)-\left\{2\right\}$ /

Differentiate $f\left(x\right)=\frac{x}{x-2}$ :

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

Since, $f\text{'}\left(x\right)$ is not defined for

  $\begin{array}{l}x-2=0\\ x=2\end{array}$ .

Therefore, the domain of $f\text{'}\left(x\right)$ is also $\left(-\infty ,\infty \right)-\left\{2\right\}$ .

Hence,

The domain of $f$ and $f\text{'}$ is $\left(-\infty ,\infty \right)-\left\{2\right\}$ .