Question

In Problems 63–66, (a) graph each function by hand, and (b) solve $f\left(x\right)\ge 0.$

$f\left(x\right)=\frac{x^3+2x^2-11x-12}{x^2-x-6}$

Short answer

Part a. The graph of the function is

Part b. The solution of$f\left(x\right)\ge 0$is$\left(-4\le x<-2and-1\le x<\infty \right)-\left\{3\right\}.$

Step-by-step solution

Part (a) Step 1. Given Information 

The given function is$f\left(x\right)=\frac{x^3+2x^2-11x-12}{x^2-x-6}$

We have to graph the function.

Part (a) Step 2. Sketch the graph of the function

The graph of the function is

Part (b) Step 1. Given Information

The given function is $f\left(x\right)=\frac{x^3+2x^2-11x-12}{x^2-x-6}$

We have to solve for$f\left(x\right)\ge 0$

Part (b) Step 2. Solving for f(x)≥0

So First, case:

$$\begin{gathered} \frac{x^3+2x^2-11x-12}{x^2-x-6}\ge 0 \\ {x}^3+2x^2-11x-12\ge 0and{x}^2-x-6>0 \\ \left(x+1\right)\left(x+4\right)\left(x-3\right)\ge 0\left(x-3\right)\left(x+2\right)>0 \\ -4\le x\le -1andx\ge 3x>3andx<-2 \\ [Commonsolution]x>3and-4\le x<-2 \end{gathered}$$

Second case:

$$\begin{gathered} \frac{x^3+2x^2-11x-12}{x^2-x-6}\ge 0 \\ {x}^3+2x^2-11x-12\ge 0and{x}^2-x-6>0 \\ \left(x+1\right)\left(x+4\right)\left(x-3\right)\ge 0\left(x-3\right)\left(x+2\right)>0 \\ -1\le x\le 3andx\le -4-2<x<3 \\ [Commonsolution]-1\le x<3 \end{gathered}$$

Thus, the solution is$\left(-4\le x<-2and-1\le x<\infty \right)-\left\{3\right\}.$