the function (f(x)= - 4x^{3}-18x^{2}+120x - 8) is increasing on the interval ( , ). it is decreasing on the…

the function (f(x)= - 4x^{3}-18x^{2}+120x - 8) is increasing on the interval ( , ). it is decreasing on the interval ((-infty,\text{ })) and the interval ((\text{ },infty)). the function has a relative maximum at (\text{ }.) attempt 1: 10 attempts remaining. submit answer next item

the function (f(x)= - 4x^{3}-18x^{2}+120x - 8) is increasing on the interval ( , ). it is decreasing on the interval ((-infty,\text{ })) and the interval ((\text{ },infty)). the function has a relative maximum at (\text{ }.) attempt 1: 10 attempts remaining. submit answer next item

Answer

Explanation:

Step1: Find the derivative of the function

Given (f(x)=- 4x^{3}-18x^{2}+120x - 8), using the power - rule ((x^n)^\prime=nx^{n - 1}), we have (f^\prime(x)=-12x^{2}-36x + 120=-12(x^{2}+3x - 10)).

Step2: Factor the derivative

Factor (x^{2}+3x - 10=(x + 5)(x - 2)), so (f^\prime(x)=-12(x + 5)(x - 2)).

Step3: Find the critical points

Set (f^\prime(x)=0), then (-12(x + 5)(x - 2)=0). Solving for (x), we get (x=-5) and (x = 2).

Step4: Determine the sign of the derivative on intervals

We consider the intervals ((-\infty,-5)), ((-5,2)) and ((2,\infty)).

  • For (x\in(-\infty,-5)), let (x=-6), then (f^\prime(-6)=-12(-6 + 5)(-6 - 2)=-12\times(-1)\times(-8)=-96<0), so the function is decreasing on ((-\infty,-5)).
  • For (x\in(-5,2)), let (x = 0), then (f^\prime(0)=-12(0 + 5)(0 - 2)=-12\times5\times(-2)=120>0), so the function is increasing on ((-5,2)).
  • For (x\in(2,\infty)), let (x = 3), then (f^\prime(3)=-12(3 + 5)(3 - 2)=-12\times8\times1=-96<0), so the function is decreasing on ((2,\infty)).
  • Since the function changes from increasing to decreasing at (x = 2), it has a relative maximum at (x = 2).

Answer:

The function has a relative maximum at (x = 2). It is decreasing on the interval ((-\infty,-5)) and the interval ((2,\infty)). It is increasing on the interval ((-5,2)).