in exercises 10-13, graph the function. identify the x - intercepts and the points where the local maximums…

Explanation:

Step1: Find the derivative

The derivative of $f(x)=0.5x^{3}-3x^{2}+1.5$ is $f^\prime(x)=1.5x^{2}-6x = 1.5x(x - 4)$.

Step2: Find the critical - points

Set $f^\prime(x)=0$. Then $1.5x(x - 4)=0$. The solutions are $x = 0$ and $x = 4$.

Step3: Determine the intervals of increase and decrease

Test the intervals $(-\infty,0)$, $(0,4)$ and $(4,\infty)$. For $x\in(-\infty,0)$, let $x=-1$. Then $f^\prime(-1)=1.5\times(-1)\times(-1 - 4)=7.5>0$, so the function is increasing on $(-\infty,0)$. For $x\in(0,4)$, let $x = 1$. Then $f^\prime(1)=1.5\times1\times(1 - 4)=-4.5<0$, so the function is decreasing on $(0,4)$. For $x\in(4,\infty)$, let $x = 5$. Then $f^\prime(5)=1.5\times5\times(5 - 4)=7.5>0$, so the function is increasing on $(4,\infty)$.

Step4: Find local maxima and minima

Since the function changes from increasing to decreasing at $x = 0$, $f(0)=0.5\times0^{3}-3\times0^{2}+1.5 = 1.5$ is a local maximum. Since the function changes from decreasing to increasing at $x = 4$, $f(4)=0.5\times4^{3}-3\times4^{2}+1.5=0.5\times64-3\times16 + 1.5=32-48 + 1.5=-14.5$ is a local minimum.

Step5: Find the x - intercepts

Set $f(x)=0.5x^{3}-3x^{2}+1.5 = 0$. Using a numerical method (such as Newton - Raphson method or a graphing utility), we find the x - intercepts. Let $y = 0.5x^{3}-3x^{2}+1.5$. The x - intercepts are approximately $x=-0.69,0.77,5.92$.

Answer:

x - intercepts: $-0.69$, $0.77$, $5.92$ local maximum: $(0,1.5)$ local minimum: $(4,-14.5)$ increasing: $x<0$ and $x > 4$ decreasing: $0<x<4$