Question

$f(x) = -4x^3 + 18x^2 + 48x + 6$
is increasing on the interval ( $square$ , $square$ ).
it is decreasing on the interval $(-\infty, \square)$ and the interval ( $square$ , $\infty$ ).
the function has a relative maximum at $square$.

Explanation:

Step1: Find first derivative

$f'(x) = \frac{d}{dx}(-4x^3 + 18x^2 + 48x + 6) = -12x^2 + 36x + 48$

Step2: Factor derivative

Factor out -12:
$f'(x) = -12(x^2 - 3x - 4) = -12(x-4)(x+1)$

Step3: Find critical points

Set $f'(x)=0$:
$-12(x-4)(x+1)=0$
Solutions: $x=-1, x=4$

Step4: Test intervals for increase/decrease

• For $x < -1$: $f'(x) < 0$ (function decreasing)
• For $-1 < x < 4$: $f'(x) > 0$ (function increasing)
• For $x > 4$: $f'(x) < 0$ (function decreasing)

Step5: Identify relative maximum

The function switches from increasing to decreasing at $x=4$, so this is a relative maximum.

Answer:

Increasing interval: $(-1, 4)$
Decreasing intervals: $(-\infty, -1)$ and $(4, \infty)$
Relative maximum at $x=4$