Question

the function $f(x) = -2x^3 - 3x^2 + 12x - 3$ is increasing on the interval ( , ). it is decreasing on the interval $(-\infty, )$ and the interval ( , $\infty$ ). the function has a relative maximum at .

Explanation:

Step1: Find first derivative

$f'(x) = \frac{d}{dx}(-2x^3 - 3x^2 + 12x - 3) = -6x^2 -6x +12$

Step2: Find critical points

Set $f'(x)=0$:
$-6x^2 -6x +12=0$
Divide by $-6$: $x^2 +x -2=0$
Factor: $(x+2)(x-1)=0$
Solutions: $x=-2, x=1$

Step3: Test intervals for increase/decrease

• For $(-\infty, -2)$: pick $x=-3$, $f'(-3)=-6(9)-6(-3)+12=-54+18+12=-24<0$ (decreasing)
• For $(-2, 1)$: pick $x=0$, $f'(0)=0-0+12=12>0$ (increasing)
• For $(1, \infty)$: pick $x=2$, $f'(2)=-6(4)-6(2)+12=-24-12+12=-24<0$ (decreasing)

Step4: Identify relative maximum

The function changes from increasing to decreasing at $x=1$, so relative maximum at $x=1$.

Answer:

Increasing interval: $(-2, 1)$
Decreasing intervals: $(-\infty, -2)$ and $(1, \infty)$
Relative maximum at $x=1$