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

Explanation:

Step1: Find the derivative

The derivative of $f(x)=0.5x^{3}-3x^{2}+1.5$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=1.5x^{2}-6x=1.5x(x - 4)$.

Step2: Find the x - intercepts

Set $f(x)=0$, so $0.5x^{3}-3x^{2}+1.5 = 0$. Multiply through by 2 to get $x^{3}-6x^{2}+3=0$. This is a cubic equation. Using a graphing utility or numerical methods (such as Newton - Raphson), we find the approximate x - intercepts. Another way is to first note that by the rational root theorem, possible rational roots are factors of 3 (i.e., $\pm1,\pm3$). None of these are roots. Using a graphing calculator, the x - intercepts are approximately $x\approx0.73,5.84$.

Step3: Find critical points

Set $f'(x)=0$. Since $f'(x)=1.5x(x - 4)=0$, the critical points are $x = 0$ and $x = 4$.

Step4: Determine local maxima and minima

We use the second - derivative test. The second - derivative $f''(x)=3x-6$. For $x = 0$, $f''(0)=-6<0$, so $f(x)$ has a local maximum at $x = 0$. $f(0)=1.5$. For $x = 4$, $f''(4)=3\times4 - 6=6>0$, so $f(x)$ has a local minimum at $x = 4$. $f(4)=0.5\times4^{3}-3\times4^{2}+1.5=0.5\times64-3\times16 + 1.5=32-48 + 1.5=-14.5$.

Step5: Find intervals of increase and decrease

We consider the intervals $(-\infty,0)$, $(0,4)$ and $(4,\infty)$. For $x\in(-\infty,0)$, pick $x=-1$, then $f'(-1)=1.5\times(-1)\times(-1 - 4)=7.5>0$, so $f(x)$ is increasing on $(-\infty,0)$. For $x\in(0,4)$, pick $x = 1$, then $f'(1)=1.5\times1\times(1 - 4)=-4.5<0$, so $f(x)$ is decreasing on $(0,4)$. For $x\in(4,\infty)$, pick $x = 5$, then $f'(5)=1.5\times5\times(5 - 4)=7.5>0$, so $f(x)$ is increasing on $(4,\infty)$.

Answer:

• x - intercepts: approximately $x\approx0.73,5.84$
• Local maximum at $(0,1.5)$
• Local minimum at $(4,-14.5)$
• Increasing intervals: $(-\infty,0)$ and $(4,\infty)$
• Decreasing interval: $(0,4)$