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

use the derivative f(x)=(x - 1)(x + 2)(x + 5) to determine the local maxima and minima of f and the intervals of increase (unique). local maximum/maxima is/are at x = -2. (use a comma to separate answers as needed.) local minimum/minima is/are at x = . (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 intervals of increase (unique). local maximum/maxima is/are at x = -2. (use a comma to separate answers as needed.) local minimum/minima is/are at x = . (use a comma to separate answers as needed.)

Answer

Explanation:

Step1: Find critical points

Set (f^{\prime}(x)=(x - 1)(x + 2)(x + 5)=0). Then (x=1,x=-2,x=-5) are critical - points.

Step2: Use the sign - change test

Consider the intervals ((-\infty,-5)), ((-5,-2)), ((-2,1)) and ((1,\infty)). For (x\in(-\infty,-5)), let (x=-6), then (f^{\prime}(-6)=(-6 - 1)(-6 + 2)(-6 + 5)=(-7)\times(-4)\times(-1)=-28<0), so (f(x)) is decreasing on ((-\infty,-5)). For (x\in(-5,-2)), let (x=-3), then (f^{\prime}(-3)=(-3 - 1)(-3 + 2)(-3 + 5)=(-4)\times(-1)\times2 = 8>0), so (f(x)) is increasing on ((-5,-2)). For (x\in(-2,1)), let (x = 0), then (f^{\prime}(0)=(0 - 1)(0 + 2)(0 + 5)=(-1)\times2\times5=-10<0), so (f(x)) is decreasing on ((-2,1)). For (x\in(1,\infty)), let (x = 2), then (f^{\prime}(2)=(2 - 1)(2 + 2)(2 + 5)=1\times4\times7 = 28>0), so (f(x)) is increasing on ((1,\infty)). Since (f(x)) changes from decreasing to increasing at (x=-5) and (x = 1), they are local minima. Since (f(x)) changes from increasing to decreasing at (x=-2), it is a local maximum.

Answer:

local minimum/minima is/are at (x=-5,1)