Chapter 5, Problem 62EP

(a)

To determine

To compute $g\left(4\right)$ and $g\left(-2\right)$ .

Answer to Problem 62EP

  $g\left(-2\right)=-\frac{9}{2}$

And

  $g\left(4\right)=2$

Explanation of Solution

Given: Graph is

  

and

  $g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

Calculation:

  $\because g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

  $\begin{array}{l}\therefore g\left(4\right)={\displaystyle \underset{1}{\overset{4}{\int }}f\left(t\right)}dt\\ ={\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)}dt+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)}dt+{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)}dt\\ =2+\left(\frac{1}{2}\times 1\times 1\right)-\left(\frac{1}{2}\times 1\times 1\right)\\ =2\end{array}$

and

  $\begin{array}{l}g\left(-2\right)={\displaystyle \underset{1}{\overset{-2}{\int }}f\left(t\right)}dt\\ =-{\displaystyle \underset{-2}{\overset{1}{\int }}f\left(t\right)}dt\\ =-\left(\frac{1}{2}\times 3\times 3\right)\\ =-\frac{9}{2}\end{array}$

(b)

To determine

To find the instantaneous rate of change of g with respect to $x$ at $x=2$ .

Answer to Problem 62EP

  $g^{\prime }\left(2\right)=1$

Explanation of Solution

Given:

  $g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

Calculation:

Since, $g^{\prime }\left(x\right)=f\left(x\right)$ is given

So instantaneous rate of change of g with respect to $x$ at $x=2$ is

  $\begin{array}{l}{g}^{\prime }\left(2\right)=f\left(2\right)\\ g^{\prime }\left(2\right)=1\end{array}$

(c)

To determine

To find the absolute minimum value of g on the interval $\left[-24\right]$ .

Answer to Problem 62EP

Absolute minimum value of g on the interval $\left[-24\right]$ is $\left(-\frac{9}{2}\right)$ at $x=-2$ .

Explanation of Solution

Given:

  $g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

Calculation:

  $\because g\left(x\right)={\displaystyle \underset{1}{\overset{x}{\int }}f\left(t\right)}dt$

  $\therefore g\left(-2\right)={\displaystyle \underset{1}{\overset{-2}{\int }}f\left(t\right)}dt$

  $\begin{array}{l}=-{\displaystyle \underset{-2}{\overset{1}{\int }}f\left(t\right)}dt\\ =-\frac{1}{2}\times 3\times 3\\ =-\frac{9}{2}\end{array}$

  $\begin{array}{l}g\left(-1\right)={\displaystyle \underset{1}{\overset{-1}{\int }}f\left(t\right)}dt\\ =-{\displaystyle \underset{-1}{\overset{1}{\int }}f\left(t\right)}dt\\ =-\left[\left(1\times 2\right)+\left(\frac{1}{2}\times 2\times 2\right)\right]\\ =-4\end{array}$

  $\begin{array}{l}g\left(0\right)={\displaystyle \underset{1}{\overset{0}{\int }}f\left(t\right)}dt\\ =-{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)}dt\\ =-\left[\left(1\times 2\right)+\frac{1}{2}\times 1\times 1\right]\\ =-\frac{5}{2}\end{array}$

  $\begin{array}{l}g\left(1\right)={\displaystyle \underset{1}{\overset{1}{\int }}f\left(t\right)}dt\\ =0\\ g\left(3\right)={\displaystyle \underset{1}{\overset{3}{\int }}f\left(t\right)}dt\\ ={\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)}dt+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)}dt\\ =2+\left(\frac{1}{2}\times 1\times 1\right)\\ =\frac{5}{2}\end{array}$

  $\begin{array}{l}g\left(4\right)={\displaystyle \underset{1}{\overset{4}{\int }}f\left(t\right)}dt\\ ={\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)}dt+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)}dt+{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)}dt\\ =2+\left(\frac{1}{2}\times 1\times 1\right)-\left(\frac{1}{2}\times 1\times 1\right)\\ =2\end{array}$

Hence absolute minimum value of g on the interval $\left[-24\right]$ is $\left(-\frac{9}{2}\right)$ at $x=-2$ .

(d)

To determine

To find the point of inflection of $g\left(x\right)$ .

Answer to Problem 62EP

  $x=1$ is the inflection point.

Explanation of Solution

Also,

  $\therefore g^{″}\left(x\right)=f^{\prime }\left(x\right)$

It is known that if $g^{″}\left(x\right)>0$ then $g\left(x\right)$ is concave up and if $g^{″}\left(x\right)<0$ then $g\left(x\right)$ is concave down.

  $\begin{array}{l}\because g^{″}\left(x\right)>0\text{on}x\in \left(-2,1\right)\\ g^{″}\left(x\right)<0\text{on}x\in \left(1,4\right)\end{array}$

Hence $x=1$ is the inflection point.