identify the inflection points and local maxima and minima of the graphed function. identify the open…

Explanation:

Step1: Find the first - derivative

Given $y = \frac{x^{3}}{3}-x^{2}-3x$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $y'=x^{2}-2x - 3$.

Step2: Find the critical points

Set $y' = 0$, so $x^{2}-2x - 3=(x - 3)(x+1)=0$. Solving gives $x=-1$ and $x = 3$.

Step3: Find the second - derivative

Differentiate $y'=x^{2}-2x - 3$ with respect to $x$. Using the power rule, $y'' = 2x-2$.

Step4: Find the inflection point

Set $y''=0$, then $2x - 2=0$, which gives $x = 1$. When $x = 1$, $y=\frac{1^{3}}{3}-1^{2}-3\times1=\frac{1}{3}-1 - 3=\frac{1 - 3-9}{3}=-\frac{11}{3}$. So the inflection point is $(1,-\frac{11}{3})$.

Step5: Determine local maxima and minima

Use the first - derivative test. For $x<-1$, let $x=-2$, then $y'=(-2)^{2}-2\times(-2)-3=4 + 4-3 = 5>0$, so the function is increasing. For $-1<x<3$, let $x = 0$, then $y'=0^{2}-2\times0 - 3=-3<0$, so the function is decreasing. For $x>3$, let $x = 4$, then $y'=4^{2}-2\times4 - 3=16-8 - 3 = 5>0$, so the function is increasing. Since the function changes from increasing to decreasing at $x=-1$, $y(-1)=\frac{(-1)^{3}}{3}-(-1)^{2}-3\times(-1)=-\frac{1}{3}-1 + 3=\frac{-1 - 3 + 9}{3}=\frac{5}{3}$ is a local maximum.

Answer:

Find the inflection point(s): A. The point(s) is/are $(1,-\frac{11}{3})$ Find each local maximum: A. There is one local maximum value of $\frac{5}{3}$ at $x=-1$