Behavior of accumulation functions

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

Looking at the graph, $f(x)>0$ when it’s above the $x$-axis on the interval

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

$f(x)<0$ when it’s below the $x$-axis on the intervals

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

$F(x)$ has a relative min when $f(x)$ changes from negative to positive. This occurs at $x=4$. To write the ordered pair, find the $y$-coordinate for this corresponding $x$-value, which would be

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

The definite integral of $f$ from $1$ to $4$ is the area of the regions formed between the graph of $f$ and the $x$-axis, which consist of:

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

Then the definite integral is the sum of those areas

$${\int }_1^{4}f(t)dt$$

$$=-\frac{1}{2}-\frac{\pi }{2}$$

And

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

The ordered pair where $F(x)$ has a relative min is $\boxed{(4,-0.071)}$.

$F(x)$ has a relative max when $f(x)$ changes from positive to negative. This occurs at $x=6.5$. To write the ordered pair, find the $y$-coordinate for this corresponding $x$-value, which would be

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

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

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

$$=1.679$$

The ordered pair where $F(x)$ has a relative max is $\boxed{(6.5,1.679)}$.

In part b), we found the relative min to be at $(4,-0.071)$ and the relative max to be at $(6.5,1.679)$.

To determine absolute extrema, we must use the extreme value theorem and test the endpoints $x=0$ and $x=8$.

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

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

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

$$=3.5$$

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

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

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

$$=-0.571$$

Then the absolute extrema are at the endpoints, with the absolute max being at

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

And the absolute min at

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

$F(x)$ is concave up when $F^{\prime \prime }(x)=f^{\prime }(x)>0$ and concave down when $f^{\prime }(x)<0$.

This means we look at where the graph of $f$ is increasing or decreasing.

$f$ is increasing on the intervals $\boxed{(0,2)\cup (3,4)\cup (4,5)}$ which is where $F(x)$ is concave up.

Note that $f^{\prime }$ is undefined at $x=2$ and $4$ and $=0$ at $x=3$ and that the open intervals do not include those points.

$f$ is decreasing on the intervals $\boxed{(2,3)\cup (6,8)}$ which is where $F(x)$ is concave down.

$F(x)$ has points of inflection where $F^{\prime \prime }(x)=f^{\prime }(x)=0$ or is undefined and changes sign.

From the graph, $f^{\prime }(x)=0$ at $x=3$ only.

$f^{\prime }(x)$ is undefined where the graph has sharp corners at $x=2,4,5,$ and $6$.

Around $x=2,f^{\prime }(x)$ changes from positive to negative, so it’s an inflection point.

The ordered pair is

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

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

$$=1.5$$

Inflection point at

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

Around $x=3,f^{\prime }(x)$ changes from negative to positive, so it’s an inflection point.

The ordered pair is

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

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

$$\approx 0.715$$

Inflection point at

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

Around $x=4,f^{\prime }(x)$ goes from positive and stays positive, so it is not an inflection point.

Around $x=5,f^{\prime }(x)$ goes from positive to $0$, so it’s not an inflection point.

Around $x=6,f^{\prime }(x)$ goes from $0$ to negative, so not an inflection point.

$h(x)$ has a relative max where it increases and then decreases. We need to find where $h^{\prime }(x)=0$ or is undefined and changes from positive to negative.

Using the FTC with the chain rule,

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

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

$h^{\prime }(x)=0$ when $2f(2x-1)=0$. Look for where $f$ crosses the $x$-axis - $f(3)$ and $f(5)$ both equal $0$.

At the input of $3,f$ changes from negative to positive, making it a min on $h(x)$.

On either side of the input $5,f$ changes from positive to negative, making it a relative max.

Since $f(5)=0$, we solve for

$$2x-1=5$$

$$x=3$$

Therefore $h^{\prime }(x)=0$ when $x=3$. To find the coordinate of the local max on $h(x)$, we find $h(3)$ which is in the next part.

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

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

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
  • $A=\frac{1}{2}(1)(2)=1$

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

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

Since $h^{\prime }(x)=2f(2x-1)$ by the FTC with the chain rule, then

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

$$=\boxed{0}$$

This is consistent with the relative extrema at $x=3$ on $h$.

Since $h^{\prime }(x)=2f(2x-1)$, taking the derivative again gives

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

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

Then

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

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

$f^{\prime }(5)$ is the slope of the tangent line at input $5$ on the graph. The tangent line is just the line that’s between $(4,1)$ and $(7,-2)$ itself, which has slope

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

$$=-1$$

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

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