use the derivative f(x)=(x - 1)(x + 2)(x + 5) to determine the local maxima and minima of f and the interval…

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

Choose test - points in the intervals $(-\infty,-5),(-5,-2),(-2,1),(1,\infty)$. For the interval $(-\infty,-5)$, let $x=-6$. Then $f'(-6)=(-6 - 1)(-6 + 2)(-6 + 5)=(-7)\times(-4)\times(-1)=-28<0$. For the interval $(-5,-2)$, let $x=-3$. Then $f'(-3)=(-3 - 1)(-3 + 2)(-3 + 5)=(-4)\times(-1)\times2 = 8>0$. For the interval $(-2,1)$, let $x=0$. Then $f'(0)=(0 - 1)(0 + 2)(0 + 5)=(-1)\times2\times5=-10<0$. For the interval $(1,\infty)$, let $x = 2$. Then $f'(2)=(2 - 1)(2 + 2)(2 + 5)=1\times4\times7 = 28>0$.

Step3: Find intervals of increase

The function $f(x)$ is increasing when $f'(x)>0$. So the intervals of increase are $(-5,-2),(1,\infty)$.

Answer:

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