Question

For the function f(x) = x² + 4x - 12 solve the following.

Answer 1

$\begin{gathered} \text{Given} \\ f(x)=x^2+4x-12 \\ \text{where} \\ f(x)\ge0 \end{gathered}$

Substitute the given equation for f(x)

$\begin{gathered} f(x)\ge0 \\ x^2+4x-12\ge0 \end{gathered}$

Factor the left side of the inequality

$\begin{gathered} (x-2)(x+6)\ge0 \\ \\ \text{separate each factor},\text{ and equate to zero to determine the intervals} \\ x-2=0 \\ x-2+2=0+2 \\ x\cancel{-2+2}=2 \\ x=2 \\ \\ x+6=0 \\ x+6-6=0-6 \\ x\cancel{+6-6}=-6 \\ x=-6 \end{gathered}$

Test for each interval

$\begin{gathered} \text{IF }x\le-6,\text{ substitute }x=-7 \\ x^2+4x-12\ge0 \\ (-7)^2+4(-7)-12\ge0 \\ 49-28-12\ge0 \\ 9\ge0 \\ x\le-6\text{ works in the original inequality} \\ \\ \text{IF }-6\le x\le2,\text{ substitute }x=0 \\ 0^2+4(0)-12\ge0 \\ -12\ngeq0 \\ -6\le x\le2\text{ does not work} \\ \\ \text{IF }x\ge2,\text{ substitute }x=3 \\ x^2+4x-12\ge0 \\ 3^2+4(3)-12\ge0 \\ 9+12-12\ge0 \\ 9\ge0 \\ x\ge2\text{ works} \end{gathered}$

Therefore, the solution to the inequality is

$(-\infty,-6\rbrack\cup\lbrack2,\infty)$