Use the graph to solve the inequality. $x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12 \geq 0$
A) $( - 2,2 ) \cup ( 3 , \infty )$
B) $[ - \infty , 3 ) \cup ( 2,3 ]$
C) $( - 2,2 ]$
D) $[ - 2,2 ] \cup [ 3 , \infty )$
E) no solution

Question

Use the graph to solve the inequality. $x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12 \geq 0$ 
A) $( - 2,2 ) \cup ( 3 , \infty )$ 
B) $[ - \infty , 3 ) \cup ( 2,3 ]$ 
C) $( - 2,2 ]$ 
D) $[ - 2,2 ] \cup [ 3 , \infty )$ 
E) no solution

Quizplus · Accepted Answer

To solve the inequality, we first need to find the $x$-values where the function $f(x) = x^3-3x^2-4x+12$ is equal to zero (i.e. the roots of the function). These $x$-values divide the number line into intervals, and we must test a point in each interval to determine whether $f(x)$ is positive or negative in that interval. Using synthetic division or polynomial long division, we can find that the function can be factored as: $$f(x) = (x-2)(x+1)^2$$ Therefore, the roots are $x=2$ and $x=-1$, and the number line is divided into three intervals: $(-\infty,-1)$, $(-1,2)$, and $(2,\infty)$. We can test a point in each interval to determine whether $f(x)$ is positive or negative in that interval. For example, if we test $x=0$ in $f(x)$, we get: $$f(0) = 12 > 0$$ Therefore, $f(x)$ is positive for any $x$ in the interval $(-1,2)$. Using similar reasoning, we can construct the following sign chart: \begin{array}{c|ccc|c} & x<-1 & -12 & x \ \hline x^3-3x^2-4x+12 & - & + & - & \ \end{array} From the sign chart, we see that $f(x)$ is non-negative (i.e. greater than or equal to zero) for $x\in[-1,2]$ and $x\geq 3$. Therefore, the solution to the inequality $f(x)\geq 0$ is: $$x\in[-1,2]\cup[3,\infty)$$ The correct answer is D, $[ - 2,2 ] \cup [ 3 , \infty )$.

Use the Graph to Solve the Inequality x3−3x2−4x+12≥0x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12 \geq 0x3−3x2−4x+12≥0

Use the Graph to Solve the Inequality $x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12 \geq 0$