3. a continuous function g is defined on the closed interval -8 ≤ x ≤ 6. the graph of g, shown above…

Explanation:

Step1: Recall the fundamental theorem of calculus

If $f(x)=\int_{a}^{x}g(t)dt$, then $f'(x) = g(x)$. Critical - points of $f$ occur where $f'(x)=0$, so we need to find where $g(x) = 0$. From the graph of $g$, $g(x)=0$ at $x=- 4$ and $x = 3$. To classify the critical - points, we use the first - derivative test. For $x\in(-8,-4)$, $g(x)<0$, so $f'(x)<0$ and $f$ is decreasing. For $x\in(-4,3)$, $g(x)>0$, so $f'(x)>0$ and $f$ is increasing. For $x\in(3,6)$, $g(x)<0$, so $f'(x)<0$ and $f$ is decreasing. So $x = - 4$ is a local minimum and $x = 3$ is a local maximum.

Step2: Calculate $f(0)$

$f(0)=\int_{-8}^{0}g(t)dt$. We split the integral into parts based on the behavior of $g$. The integral from $x=-8$ to $x = - 4$: The area of the triangle with base $4$ and height $4$ (from $(-8,-2)$ to $(-4,2)$) is $A_1=\frac{1}{2}\times4\times4 = 8$. The integral from $x=-4$ to $x = 0$: The area of the quarter - circle with radius $4$ centered at $(0,2)$ is $A_2=\frac{1}{4}\pi r^{2}=\frac{1}{4}\pi\times4^{2}=4\pi$. So $f(0)=8 + 4\pi$.

Step3: Apply L'Hopital's rule

We have $\lim_{x\rightarrow - 4}\frac{f(x)}{x^{2}+4x}$. Since $f(-4)=\int_{-8}^{-4}g(t)dt$ and $x^{2}+4x$ evaluated at $x=-4$ is $(-4)^{2}+4\times(-4)=0$, and by the fundamental theorem of calculus $f'(x)=g(x)$. By L'Hopital's rule, $\lim_{x\rightarrow - 4}\frac{f(x)}{x^{2}+4x}=\lim_{x\rightarrow - 4}\frac{f'(x)}{2x + 4}=\lim_{x\rightarrow - 4}\frac{g(x)}{2x + 4}$. From the graph, $g(-4)=0$. Substituting $x=-4$ into $\frac{g(x)}{2x + 4}$, we get $\frac{0}{2\times(-4)+4}=0$.

Step4: Use the quotient rule

The quotient rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^{2}}$, where $u(x)=g(x)$ and $v(x)=x^{2}+1$. So $h'(x)=\frac{g'(x)(x^{2}+1)-g(x)\times2x}{(x^{2}+1)^{2}}$. From the graph, $g(1)$: We need to find the equation of the line segment from $(0,2)$ to $(3,-3)$. The slope of the line is $m=\frac{-3 - 2}{3-0}=-\frac{5}{3}$, and the equation of the line is $y-2=-\frac{5}{3}(x - 0)$ or $y=2-\frac{5}{3}x$. When $x = 1$, $g(1)=2-\frac{5}{3}=\frac{1}{3}$. The slope of the line segment at $x = 1$ (i.e., $g'(1)$) is $-\frac{5}{3}$. Substitute $x = 1$, $g(1)=\frac{1}{3}$, and $g'(1)=-\frac{5}{3}$ into the formula for $h'(x)$: [ \begin{align*} h'(1)&=\frac{-\frac{5}{3}(1^{2}+1)-\frac{1}{3}\times2\times1}{(1^{2}+1)^{2}}\ &=\frac{-\frac{5}{3}\times2-\frac{2}{3}}{4}\ &=\frac{-\frac{10 + 2}{3}}{4}\ &=\frac{-4}{4}\ &=-1 \end{align*} ]

Answer:

(a) Critical points are $x=-4$ (local minimum) and $x = 3$ (local maximum). (b) $f(0)=8 + 4\pi$ (c) $0$ (d) $-1$