Question

the function
$f(x) = 6x^3 + 36x^2 - 576x + 6$
is decreasing on the interval ( , ).
it is increasing on the interval $(-\infty, )$
and the interval ( , $\infty$).
the function has a local maximum at .

Explanation:

Step1: Find first derivative

$f'(x) = \frac{d}{dx}(6x^3 + 36x^2 - 576x + 6) = 18x^2 + 72x - 576$

Step2: Simplify derivative

Factor out 18: $f'(x) = 18(x^2 + 4x - 32)$

Step3: Find critical points

Set $f'(x)=0$:
$x^2 + 4x - 32 = 0$
Factor: $(x+8)(x-4)=0$
Solutions: $x=-8, x=4$

Step4: Test intervals for sign

• For $x < -8$: $f'(x) > 0$ (function increasing)
• For $-8 < x < 4$: $f'(x) < 0$ (function decreasing)
• For $x > 4$: $f'(x) > 0$ (function increasing)

Step5: Identify local maximum

Function switches from increasing to decreasing at $x=-8$, so this is a local maximum.

Answer:

Decreasing interval: $(-8, 4)$
Increasing interval upper bound: $-8$
Increasing interval lower bound: $4$
Local maximum at: $x=-8$