Because

$$F^{\prime }(x)=f(x),$$

$F(x)$ is increasing where $f(x)>0$ and decreasing where $f(x)<0$.

From the graph, $f(x)>0$ (above the $x$-axis) on

$$\boxed{(4,6.5)}.$$

From the graph, $f(x)<0$ (below the $x$-axis) on

$$\boxed{(0,2)\cup (2,4)\cup (6.5,8)}.$$

A relative minimum of $F$ occurs where $F^{\prime }(x)$ changes from negative to positive. Since $F^{\prime }(x)=f(x)$, this happens where $f(x)$ changes from negative to positive.

From the graph, $f(x)$ changes from negative to positive at $x=4$, so $F$ has a relative minimum at $x=4$.

To write the ordered pair, compute the $y$-value:

$$F(4)=2+{\int }_1^{4}f(t)dt.$$

Interpret the definite integral as signed area between the graph of $f$ and the $x$-axis from $t=1$ to $t=4$. The regions are:

• A triangle with signed area $=\frac{1}{2}(1)(1)$
• A semi-circle with signed area $=\frac{1}{2}\pi (1{)}^2$

Because the graph is below the $x$-axis on this interval, these contribute negative signed area, so

$${\int }_1^{4}f(t)dt=-\frac{1}{2}-\frac{\pi }{2}.$$

Then

$$F(4)=2+\left(-\frac{1}{2}-\frac{\pi }{2}\right)\approx -\mathrm{0.071.}$$

So the ordered pair for the relative minimum is

$$\boxed{(4,-0.071)}.$$

A relative maximum of $F$ occurs where $f(x)$ changes from positive to negative.

From the graph, this occurs at $x=6.5$, so $F$ has a relative maximum at $x=6.5$.

Compute the $y$-value:

$$F(6.5)=2+{\int }_1^{6.5}f(t)dt$$

$$=2+\left(-\frac{1}{2}-\frac{\pi }{2}+\frac{1}{2}(1)(1+2.5)\right)$$

$$\approx 2+(-0.321)$$

$$=\mathrm{1.679.}$$

So the ordered pair for the relative maximum is

$$\boxed{(6.5,1.679)}.$$

From part b), the relative minimum is $(4,-0.071)$ and the relative maximum is $(6.5,1.679)$.

To find absolute extrema on the interval shown, use the extreme value theorem: check the endpoints and any interior critical points. Here, we compare $F(0)$, $F(8)$, and the relative extrema values.

Compute the endpoint values.

For $x=0$:

$$F(0)=2+{\int }_1^{0}f(t)dt$$

$$=2-{\int }_0^{1}f(t)dt$$

$$=2-\left(-\frac{1}{2}(1)(2+1)\right)$$

$$=\mathrm{3.5.}$$

For $x=8$:

$$F(8)=2+{\int }_1^{8}f(t)dt$$

$$=2+{\int }_1^{6.5}f(t)dt+{\int }_{6.5}^{8}f(t)dt$$

$$=2+(-0.321)+\left(-\frac{1}{2}(1.5)(3)\right)$$

$$=-\mathrm{0.571.}$$

Comparing values, the absolute maximum occurs at $x=0$ and the absolute minimum occurs at $x=8$:

$$\boxed{(0,3.5)}$$

and

$$\boxed{(8,-0.571)}.$$

Concavity comes from the second derivative:

$$F^{\prime \prime }(x)=f^{\prime }(x).$$

So $F$ is concave up where $f^{\prime }(x)>0$ (where $f$ is increasing) and concave down where $f^{\prime }(x)<0$ (where $f$ is decreasing).

From the graph, $f$ is increasing on

$$\boxed{(0,2)\cup (3,4)\cup (4,5)},$$

so $F$ is concave up on those intervals.

Note that $f^{\prime }$ is undefined at $x=2$ and $x=4$, and $f^{\prime }(x)=0$ at $x=3$. That’s why we use open intervals.

From the graph, $f$ is decreasing on

$$\boxed{(2,3)\cup (6,8)},$$

so $F$ is concave down on those intervals.

Inflection points of $F$ occur where

$$F^{\prime \prime }(x)=f^{\prime }(x)$$

changes sign. That can happen where $f^{\prime }(x)=0$ or where $f^{\prime }(x)$ is undefined, as long as the sign of $f^{\prime }$ changes.

From the graph:

• $f^{\prime }(x)=0$ at $x=3$.
• $f^{\prime }(x)$ is undefined at sharp corners at $x=2,4,5,$ and $6$.

Check sign changes in $f^{\prime }$ (increasing vs. decreasing behavior of $f$):

• Around $x=2$, $f$ changes from increasing to decreasing, so $f^{\prime }$ changes from positive to negative. This gives an inflection point of $F$.

Compute the ordered pair:

$$F(2)=2+{\int }_1^{2}f(t)dt$$

$$=2+\left(-\frac{1}{2}(1)(1)\right)$$

$$=\mathrm{1.5.}$$

Inflection point:

$$\boxed{(2,1.5)}$$

• Around $x=3$, $f$ changes from decreasing to increasing, so $f^{\prime }$ changes from negative to positive. This gives an inflection point of $F$.

Compute the ordered pair:

$$F(3)=2+{\int }_1^{3}f(t)dt$$

$$=2+\left(-\frac{1}{2}-\frac{1}{4}\pi (1{)}^2\right)$$

$$\approx \mathrm{0.715.}$$

Inflection point:

$$\boxed{(3,0.715)}$$

• Around $x=4$, $f$ is increasing on both sides, so $f^{\prime }$ stays positive. Not an inflection point.
• Around $x=5$, $f$ goes from increasing to flat ($f^{\prime }=0$) and then continues in a way that does not create a sign change in $f^{\prime }$. Not an inflection point.
• Around $x=6$, $f$ goes from flat ($f^{\prime }=0$) to decreasing, but this does not create a sign change from negative to positive or positive to negative across $x=6$ in the way needed here. Not an inflection point.

A relative maximum of $h$ occurs where $h$ changes from increasing to decreasing. That means we look for where $h^{\prime }(x)$ changes from positive to negative.

Differentiate using the FTC and the chain rule:

$$h^{\prime }(x)=f(2x-1)\cdot \frac{d}{dx}(2x-1)$$

$$=2f(2x-1).$$

So $h^{\prime }(x)=0$ when $f(2x-1)=0$. From the graph, $f$ crosses the $x$-axis at inputs $3$ and $5$ (so $f(3)=0$ and $f(5)=0$).

Now check the sign change:

• At input $3$, $f$ changes from negative to positive, which would make $h$ change from decreasing to increasing (a relative minimum).
• At input $5$, $f$ changes from positive to negative, which would make $h$ change from increasing to decreasing (a relative maximum).

Use the input $5$ and solve for $x$:

$$2x-1=5$$

$$x=3.$$

So $h$ has a relative maximum at $x=3$. To get the ordered pair, we compute $h(3)$ in part b).

Evaluate the accumulation function:

$$h(3)={\int }_0^{2(3)-1}f(t)dt$$

$$={\int }_0^{5}f(t)dt.$$

From $0$ to $5$, the regions consist of:

• A trapezoid with signed area
  • $A=-\frac{1}{2}(1+3)=-2$
• A triangle with signed area
  • $A=\frac{1}{2}(1)(2)=1$

Add the signed areas:

$$h(3)=-2+1=\boxed{-1}.$$

Combining with part a), $h(x)$ has a relative max at $(3,-1)$.

From part a),

$$h^{\prime }(x)=2f(2x-1).$$

So

$$h^{\prime }(3)=2f(5)=\boxed{0}.$$

This matches the fact that $x=3$ is a relative extremum of $h$.

Differentiate $h^{\prime }(x)$ again. Since

$$h^{\prime }(x)=2f(2x-1),$$

we get

$$h^{\prime \prime }(x)=2f^{\prime }(2x-1)\cdot \frac{d}{dx}(2x-1)$$

$$=4f^{\prime }(2x-1).$$

Then

$$h^{\prime \prime }(3)=4f^{\prime }(2(3)-1)$$

$$=4f^{\prime }(5).$$

$f^{\prime }(5)$ is the slope of the graph of $f$ at input $5$. At $x=5$, the graph lies on the line segment between $(4,1)$ and $(7,-2)$, so the slope is

$$\frac{1-(-2)}{4-7}=-1.$$

Therefore $f^{\prime }(5)=-1$, and

$$\boxed{h^{\prime \prime }(3)=-4}.$$