Chapter 4.3, Problem 5QR

To determine

To calculate : The domain of $f$ and $f^{\prime }$ for the function $f\left(x\right)=\frac{x}{x-2}$ .

Answer to Problem 5QR

The domains of both $f$ and $f^{\prime }$ for the function $f\left(x\right)=\frac{x}{x-2}$ are $\left(-\infty ,2\right)\cup \left(2,\infty \right)$respectively.

Explanation of Solution

Given information : The function $f\left(x\right)=\frac{x}{x-2}$ .

Formula used : Use the quotient rule to calculate the differentiation $f^{\prime }$ .

Calculation :

Consider the function $f\left(x\right)=\frac{x}{x-2}$ ,

The domain of the function is the set of real numbers except the values where the given function is undefined.

The rational function is defined for all the values of real numbers except where the denominator is $0$ .

The denominator will be $0$ if $x-2=0$ that is $x=2$ .

So, the domain of $f\left(x\right)=\frac{x}{x-2}$ is set of real numbers except $2$ that is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$ .

Now, to find the derivative of given function use the quotient rule of differentiation.

The value of $f^{\prime }$ for the function $f\left(x\right)=\frac{x}{x-2}$ is calculated as follows.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{\left(x-2\right)\frac{d}{dx}2-2\frac{d}{dx}\left(x-2\right)}{{\left(x-2\right)}^2}\\ =\frac{\left(x-2\right)\left(0\right)-2\left(1\right)}{{\left(x-2\right)}^2}\\ =\frac{-2}{{\left(x-2\right)}^2}\end{array}$

The function is defined for all the values except where the denominator is 0.

The denominator will be $0$ if $x-2=0$ that is $x=2$ .

The derivative is defined for all real values except $2$ that is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$ .

Thus, the domain of both $f$ is and $f^{\prime }$ for the function $f\left(x\right)=\frac{x}{x-2}$ is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$ .