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

Step2: Test intervals

We have four intervals to test: $(-\infty,-5),(-5,-2),(-2,1),(1,\infty)$. 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 $(1,\infty)$.

Step3: Determine local extrema

Since $f(x)$ changes from decreasing to increasing at $x=-5$, $f(-5)$ is a local minimum. Since $f(x)$ changes from increasing to decreasing at $x=-2$, $f(-2)$ is a local maximum. Since $f(x)$ changes from decreasing to increasing at $x = 1$, $f(1)$ is a local minimum.

A possible graph of $f$ has a local minimum at $x=-5$, a local maximum at $x=-2$, and a local minimum at $x = 1$. Looking at the graphs:

• In graph A, we can see a local minimum, then a local maximum, then a local minimum which is consistent with our analysis.
• In graph B, the order of extrema is not correct.
• In graph C, the order of extrema is not correct.

Answer:

A.