the function (f(x)=2x^{3}+12x^{2}-192x + 7) is decreasing on the interval ( ). it is increasing on the…

Explanation:

Step1: Find the derivative of the function

Given (f(x)=2x^{3}+12x^{2}-192x + 7), using the power - rule ((x^n)^\prime=nx^{n - 1}), we have (f^\prime(x)=6x^{2}+24x-192 = 6(x^{2}+4x - 32)).

Step2: Factor the derivative

Factor (x^{2}+4x - 32=(x + 8)(x - 4)), so (f^\prime(x)=6(x + 8)(x - 4)).

Step3: Find the critical points

Set (f^\prime(x)=0), then (6(x + 8)(x - 4)=0). The critical points are (x=-8) and (x = 4).

Step4: Determine the sign of the derivative in intervals

We consider the intervals ((-\infty,-8)), ((-8,4)) and ((4,\infty)).

• For (x\in(-\infty,-8)), let (x=-9), then (f^\prime(-9)=6(-9 + 8)(-9 - 4)=6\times(-1)\times(-13)=78>0), so the function is increasing on ((-\infty,-8)).
• For (x\in(-8,4)), let (x = 0), then (f^\prime(0)=6(0 + 8)(0 - 4)=6\times8\times(-4)=-192<0), so the function is decreasing on ((-8,4)).
• For (x\in(4,\infty)), let (x = 5), then (f^\prime(5)=6(5 + 8)(5 - 4)=6\times13\times1 = 78>0), so the function is increasing on ((4,\infty)). Since the function changes from increasing to decreasing at (x=-8), it has a local maximum at (x=-8).

Answer:

The function is decreasing on the interval ((-8,4)), increasing on the intervals ((-\infty,-8)) and ((4,\infty)), and has a local maximum at (x=-8).