Chapter 2.2, Problem 11Q

To determine

The $x-\text{intercept}$ of the graph of the function and decide where the function is increasing and decreasing.

Answer to Problem 11Q

  $x=5$ & $x=-1$ are the required $x-\text{intercept}$ of the function $g\left(x\right)=\frac{1}{2}\left(x-5\right)\left(x+1\right)$ and the function is increasing in $\left[2,\infty \right)$ and decreasing in $\left(-\infty ,2\right]$ .

Explanation of Solution

Given information : Given function is $g\left(x\right)=\frac{1}{2}\left(x-5\right)\left(x+1\right)$ .

To find the $x-\text{intercept}$ of the function $g\left(x\right)=\frac{1}{2}\left(x-5\right)\left(x+1\right)$ , we need to solve the equation $g\left(x\right)=0$ .

Therefore,

  $\begin{array}{c}-\frac{1}{2}\left(x-5\right)\left(x+1\right)=0\\ \left(x-5\right)\left(x+1\right)=0\\ x=5,-1\end{array}$ .

Thus, $x=5$ & $x=-1$ are the required $x-\text{intercept}$ of the function $g\left(x\right)=\frac{1}{2}\left(x-5\right)\left(x+1\right)$ .

To decide where the function is increasing and decreasing, we need to plot the graph of $g\left(x\right)=\frac{1}{2}\left(x-5\right)\left(x+1\right)$ .

  

Clearly, from the above graph, the function is increasing in $\left[2,\infty \right)$ and decreasing in $\left(-\infty ,2\right]$ .