Question

$f(x) = 4x^3 - 6x^2 - 240x - 6$
is decreasing on the interval ($\boldsymbol{square}$, $\boldsymbol{square}$).
it is increasing on the interval $(-\infty, \boldsymbol{square})$ and the interval $(\boldsymbol{square}, \infty)$.
the function has a local maximum at $\boldsymbol{square}$.

Explanation:

Step1: Find first derivative

$f'(x) = \frac{d}{dx}(4x^3 - 6x^2 - 240x - 6) = 12x^2 - 12x - 240$

Step2: Simplify derivative

Factor out 12:
$12(x^2 - x - 20) = 12(x-5)(x+4)$

Step3: Find critical points

Set $f'(x)=0$:
$12(x-5)(x+4)=0 \implies x=5, x=-4$

Step4: Test intervals for increase/decrease

• For $x < -4$, $f'(x) > 0$ (function increasing)
• For $-4 < x < 5$, $f'(x) < 0$ (function decreasing)
• For $x > 5$, $f'(x) > 0$ (function increasing)

Step5: Identify local maximum

Function switches from increasing to decreasing at $x=-4$, so this is a local maximum.

Answer:

Decreasing interval: $(-4, 5)$
Increasing interval endpoint: $-4$; Increasing interval endpoint: $5$
Local maximum at: $x=-4$