Question

the function

f(x)=4x^{3}-6x^{2}-144x + 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 the derivative

$f'(x)=12x^{2}-12x - 144$

Step2: Set the derivative equal to zero

$12x^{2}-12x - 144 = 0$. Divide through by 12: $x^{2}-x - 12=0$. Factor: $(x - 4)(x+ 3)=0$. So $x=-3$ or $x = 4$.

Step3: Test intervals

Choose test - points in the intervals $(-\infty,-3)$, $(-3,4)$ and $(4,\infty)$. Let's choose $x=-4$, $x = 0$ and $x = 5$.
$f'(-4)=12\times(-4)^{2}-12\times(-4)-144=192 + 48-144 = 96>0$.
$f'(0)=12\times0^{2}-12\times0 - 144=-144<0$.
$f'(5)=12\times5^{2}-12\times5 - 144=300-60 - 144 = 96>0$.
The function is decreasing when $f'(x)<0$, so it is decreasing on the interval $(-3,4)$. It is increasing when $f'(x)>0$, so it is increasing on $(-\infty,-3)$ and $(4,\infty)$.
A local maximum occurs where the function changes from increasing to decreasing, so the local maximum occurs at $x=-3$.

Answer:

Decreasing interval: $(-3,4)$
Increasing intervals: $(-\infty,-3)$, $(4,\infty)$
Local maximum at: $x=-3$