the function\n\n$f(x)=2x^{3}+12x^{2}-192x + 7$\n\nis decreasing on the interval ( -8, 4 ).\nit is increasing…

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 $x=-8$.