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 and d not unique). the local maximum/maxima 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 and d not unique). the local maximum/maxima 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: Create a sign - chart

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

Step3: Determine local extrema

Since $f^{\prime}(x)$ changes sign from negative to positive at $x=-5$ and $x = 1$, $f(x)$ has local minima at $x=-5$ and $x = 1$. Since $f^{\prime}(x)$ changes sign from positive to negative at $x=-2$, $f(x)$ has a local maximum at $x=-2$.

Answer:

$-2$