use the derivative f(x)=(x - 1)(x + 2)(x + 5) to determine the local maxima and minima of f (not unique)…

use the derivative f(x)=(x - 1)(x + 2)(x + 5) to determine the local maxima and minima of f (not unique). the local maximum/maxima is/are at x= -2. (use a comma to separate answers as needed. the local minimum/minima is/are at x= -5,1. (use a comma to separate answers as needed. the interval(s) of increase is(are) (-5, -2),(1,∞). (type your answer in interval notation. use a comma to separate answers as needed.) the interval(s) of decrease is(are) . (type your answer in interval notation. use a comma to separate answers as needed.)

use the derivative f(x)=(x - 1)(x + 2)(x + 5) to determine the local maxima and minima of f (not unique). the local maximum/maxima is/are at x= -2. (use a comma to separate answers as needed. the local minimum/minima is/are at x= -5,1. (use a comma to separate answers as needed. the interval(s) of increase is(are) (-5, -2),(1,∞). (type your answer in interval notation. use a comma to separate answers as needed.) the interval(s) of decrease is(are) . (type your answer in interval notation. use a comma to separate answers as needed.)

Answer

Explanation:

Step1: Find critical points

Set $f'(x)=(x - 1)(x + 2)(x + 5)=0$. The critical points are $x=-5,x=-2,x = 1$.

Step2: Determine sign of derivative in intervals

  • For $x<-5$, let $x=-6$, then $f'(-6)=(-6 - 1)(-6 + 2)(-6 + 5)=(-7)\times(-4)\times(-1)<0$.
  • For $-5<x<-2$, let $x=-3$, then $f'(-3)=(-3 - 1)(-3 + 2)(-3 + 5)=(-4)\times(-1)\times2>0$.
  • For $-2<x<1$, let $x = 0$, then $f'(0)=(0 - 1)(0 + 2)(0 + 5)=(-1)\times2\times5<0$.
  • For $x>1$, let $x = 2$, then $f'(2)=(2 - 1)(2 + 2)(2 + 5)=1\times4\times7>0$.

Step3: Find intervals of decrease

Since $f'(x)<0$ when $x<-5$ and $-2<x<1$, the function is decreasing on the intervals $(-\infty,-5)$ and $(-2,1)$.

Answer:

$(-\infty,-5),(-2,1)$