Chapter 4.6, Problem 65AYU

In Problems 63-66, (a) graph each function by hand, and (b) solve .

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

To determine

To find:

(a) Graph of the function by hand.

Answer to Problem 65AYU

$\left\{x|-4 \le x < -2, -1 \le x < 3, 3 < x < \infty \right\}$

Explanation of Solution

Given:

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

(a)

To determine

To find:

(b) Solve $f\left(x\right) \ge 0$.

Answer to Problem 65AYU

$\left\{x|-4 \le x < -2, -1 \le x < 3, 3 < x < \infty \right\}$

Explanation of Solution

Given:

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

(b) The zero of $f\left(x\right) = \frac{x^3 + 2x^2 - 11x - 12}{x^2 - x - 6} = \frac{\left(x + 1\right)\left(x + 4\right)\left(x - 3\right)}{\left(x - 3\right)\left(x + 2\right)} = \frac{\left(x + 1\right)\left(x + 4\right)}{\left(x + 2\right)}$ is $x = -1, -4, 3$. Also function not defined at $x = -2, 3$.

Use the zero and the undefined value to separate the real number line into intervals:

$\left(-\infty ,-4\right],\left[-4,-2\right),\left(-2,-1\right],\left[-1,3\right),\left(3,\infty \right)$

Select a test number in each interval found in above and evaluate $f\left(x\right) = \frac{\left(x + 1\right)\left(x + 4\right)}{\left(x + 2\right)}$ each number to determine if $f\left(x\right)$ is positive or negative.

Interval	$\left(-\infty , -4\right]$	$\left[-4, -2\right)$	$\left(-2, -1\right]$	$\left[-1, 3\right)$	$\left(3, \infty \right)$
Number chosen	$-5$	$-3$	$-1.5$	$2$	$2$
Value of $f$	$-\frac{4}{3}$	$2$	$-2.5$	$\frac{9}{2}$	$\frac{20}{3}$
Conclusion	Negative	Positive	Negative	Positive	Positive

Since we want to know where $f\left(x\right)$ is positive and zero, conclude that $f\left(x\right) \ge 0$. The solution set is $\left\{x|-4 \le x < -2, -1 \le x < 3, 3 < x < \infty \right\}$ or, using interval notation $\left[-4, -2\right){{\displaystyle \cup }}^{\text{}}\left[-1, 3\right){{\displaystyle \cup }}^{\text{}}\left(3, \infty \right)$.