this question has two parts. first, answer part a. then, answer part b. part a on a separate sheet of paper…

Explanation:

Step1: Find the derivative

The derivative of $f(x)=2x^{3}-3x^{2}+2$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=6x^{2}-6x=6x(x - 1)$.

Step2: Find the critical points

Set $f'(x)=0$. So, $6x(x - 1)=0$. Solving this equation gives $x = 0$ and $x = 1$ as the critical points.

Step3: Analyze the sign of the derivative

We consider the intervals $(-\infty,0)$, $(0,1)$ and $(1,\infty)$. For $x\in(-\infty,0)$, let $x=-1$, then $f'(-1)=6\times(-1)\times(-2)=12>0$, so $f(x)$ is increasing on $(-\infty,0)$. For $x\in(0,1)$, let $x=\frac{1}{2}$, then $f'(\frac{1}{2})=6\times\frac{1}{2}\times(-\frac{1}{2})=-\frac{3}{2}<0$, so $f(x)$ is decreasing on $(0,1)$. For $x\in(1,\infty)$, let $x = 2$, then $f'(2)=6\times2\times1 = 12>0$, so $f(x)$ is increasing on $(1,\infty)$. So, $f(x)$ has a relative maximum at $x = 0$ and a relative minimum at $x = 1$.

Step4: Analyze the zero - points

$f(-1)=2\times(-1)^{3}-3\times(-1)^{2}+2=-2 - 3+2=-3<0$ and $f(0)=2>0$. By the Intermediate Value Theorem, there is a zero between $x=-1$ and $x = 0$.

Answer:

zero between $x=-1$ and $x = 0$ relative maximum at $x = 0$ relative minimum at $x = 1$