Question

the function

$f(x)=2x^{3}+12x^{2}-192x + 7$

is decreasing on the interval ( -8, 4 ).
it is increasing on the interval ( -∞, -8 )
and the interval ( 4, ∞ ).
the function has a local maximum at -8.

Explanation:

Step1: Find the derivative

$f'(x)=6x^{2}+24x - 192$

Step2: Set the derivative equal to zero

$6x^{2}+24x - 192 = 0$
Divide through by 6: $x^{2}+4x - 32=0$
Factor: $(x + 8)(x - 4)=0$
Solve for $x$: $x=-8$ or $x = 4$

Step3: Determine the sign of the derivative

Choose test - points in the intervals $(-\infty,-8)$, $(-8,4)$ and $(4,\infty)$.
For $x=-9$ (in $(-\infty,-8)$), $f'(-9)=6\times(-9)^{2}+24\times(-9)-192=6\times81-216 - 192=486-216 - 192 = 78>0$.
For $x = 0$ (in $(-8,4)$), $f'(0)=6\times0^{2}+24\times0-192=-192<0$.
For $x = 5$ (in $(4,\infty)$), $f'(5)=6\times5^{2}+24\times5-192=6\times25 + 120-192=150+120 - 192 = 78>0$.
A function is decreasing when $f'(x)<0$ (on $(-8,4)$), increasing when $f'(x)>0$ (on $(-\infty,-8)$ and $(4,\infty)$).
The local maximum occurs where the function changes from increasing to decreasing, which is at $x=-8$.

Answer:

The function $f(x)=2x^{3}+12x^{2}-192x + 7$ is decreasing on the interval $(-8,4)$, increasing on the intervals $(-\infty,-8)$ and $(4,\infty)$, and has a local maximum at $x=-8$.