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

Explanation:

Step1: Find critical points

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

Step2: Test intervals

Consider intervals $(-\infty,-5),(-5,-2),(-2,1),(1,\infty)$. For $x\in(-\infty,-5)$, let $x=-6$, then $f'(-6)=(-6 - 1)(-6 + 2)(-6+5)=(-7)\times(-4)\times(-1)=-28<0$, so $f$ is decreasing on $(-\infty,-5)$. For $x\in(-5,-2)$, let $x = - 3$, then $f'(-3)=(-3 - 1)(-3 + 2)(-3 + 5)=(-4)\times(-1)\times2 = 8>0$, so $f$ is increasing on $(-5,-2)$. For $x\in(-2,1)$, let $x=0$, then $f'(0)=(0 - 1)(0 + 2)(0 + 5)=(-1)\times2\times5=-10<0$, so $f$ is decreasing on $(-2,1)$. For $x\in(1,\infty)$, let $x = 2$, then $f'(2)=(2 - 1)(2 + 2)(2 + 5)=1\times4\times7 = 28>0$, so $f$ is increasing on $(1,\infty)$.

Step3: Determine local extrema

Since $f$ changes from decreasing to increasing at $x=-5$ and $x = 1$, these are local minima. Since $f$ changes from increasing to decreasing at $x=-2$, it is a local maximum.

Answer:

1, - 5