Chapter 5.4, Problem 19E

a.

To determine

To find: the value of at which the local maximum and minimum values of $\mathrm{g}$ occur.

Answer to Problem 19E

The value of $\mathrm{x}$ at which the local maximum and minimum values of $\mathrm{g}$ occur at 1 and 5 and at 3 and 7 respectively.

Explanation of Solution

Given information:

The function is $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}}$ and the graph of $\mathrm{f}\left(\mathrm{t}\right)$ is shown below:

  

Concept used:

The fundamental theorem of calculus, part 1 is defined by,

If $\mathrm{f}$ is continuous on $\left[\mathrm{a},\mathrm{b}\right]$ , then the function $\mathrm{g}$ defined by,

  $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{\mathrm{a}}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)}\mathrm{d}\mathrm{t}$

  $\mathrm{a}\le \mathrm{x}\le \mathrm{b}$ is an antiderivative of $\mathrm{f}$ , that is ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ for $\mathrm{a}<\mathrm{x}<\mathrm{b}$ .

Using Part 1 of the Fundamental Theorem of Calculus,

  ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ [since, $\mathrm{f}\left(\mathrm{t}\right)$ is continuous and $0<\mathrm{x}<\mathrm{b}$ ]

Since, a function has a local maximum or local minimum at a point ${\mathrm{x}}_0$ if and only if the first derivative of a function is equal to zero at $\mathrm{x}={\mathrm{x}}_0$ .

Hence, the function $\mathrm{g}$ has a local maximum and minimum at a point where ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$ that is $\mathrm{f}\left(\mathrm{x}\right)$ is concave up and concave down respectively and touches the $\mathrm{x}$ -axis.

From the graph provided in the question, it can be observed that $\mathrm{f}$ is concave up between $0\le \mathrm{x}\le 1$ and $3\le \mathrm{x}\le 5$ and is concave down between $1\le \mathrm{x}\le 3$ and $5\le \mathrm{x}\le 7$ .

Therefore,

The value of $\mathrm{x}$ at which the local maximum and minimum values of $\mathrm{g}$ occur at 1 and 5 and at 3 and 7 respectively.

b.

To determine

To find: the value of $\mathrm{x}$ at which the absolute maximum values of $\mathrm{g}$ occur.

Answer to Problem 19E

The value of $\mathrm{x}$ at which the absolute maximum values of $\mathrm{g}$ occur at $\mathrm{x}=9$ .

Explanation of Solution

Given information:

The function is $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}}$ and the graph of $\mathrm{f}\left(\mathrm{t}\right)$ is shown below:

  

Concept used:

The fundamental theorem of calculus, part 1 is defined by,

If $\mathrm{f}$ is continuous on $\left[\mathrm{a},\mathrm{b}\right]$ , then the function $\mathrm{g}$ defined by,

  $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{\mathrm{a}}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)}\mathrm{d}\mathrm{t}$ $\mathrm{a}\le \mathrm{x}\le \mathrm{b}$ is an antiderivative of $\mathrm{f}$ , that is ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ for $\mathrm{a}<\mathrm{x}<\mathrm{b}$ .

Using Part 1 of the Fundamental Theorem of Calculus,

  ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ [since, $\mathrm{f}\left(\mathrm{t}\right)$ is continuous and $0<\mathrm{x}<\mathrm{b}$ ]

Since, a function has a local maximum or local minimum at a point ${\mathrm{x}}_0$ if and only if the first derivative of a function is equal to zero at $\mathrm{x}={\mathrm{x}}_0$ .

Hence, the function $\mathrm{g}$ has a local maximum at a point where ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$ that is $\mathrm{f}\left(\mathrm{x}\right)$ has highest concave up peak and touches the $\mathrm{x}$ -axis.

From the graph provided in the question, it can be observed that $\mathrm{f}$ has absolute maximum at $\mathrm{x}=9$ .

Therefore,

The value of $\mathrm{x}$ at which the absolute maximum values of $\mathrm{g}$ occur at $\mathrm{x}=9$ .

c.

To determine

To find: the interval at which $\mathrm{g}$ is concave downward.

Answer to Problem 19E

The interval at which the function $\mathrm{g}$ is concave down is between $1\le \mathrm{x}\le 3$ and $5\le \mathrm{x}\le 7$ .

Explanation of Solution

Given information:

The function is $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}}$ and the graph of $\mathrm{f}\left(\mathrm{t}\right)$ is shown below:

  

Concept used:

The fundamental theorem of calculus, part 1 is defined by,

If $\mathrm{f}$ is continuous on $\left[\mathrm{a},\mathrm{b}\right]$ , then the function $\mathrm{g}$ defined by,

  $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{\mathrm{a}}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)}\mathrm{d}\mathrm{t}$ $\mathrm{a}\le \mathrm{x}\le \mathrm{b}$ is an antiderivative of $\mathrm{f}$ , that is ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ for $\mathrm{a}<\mathrm{x}<\mathrm{b}$ .

Using Part 1 of the Fundamental Theorem of Calculus,

  ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ [since, $\mathrm{f}\left(\mathrm{t}\right)$ is continuous and $0<\mathrm{x}<\mathrm{b}$ ]

Since, a function has a local maximum or local minimum at a point ${\mathrm{x}}_0$ if and only if the first derivative of a function is equal to zero at $\mathrm{x}={\mathrm{x}}_0$ .

Hence, the function $\mathrm{g}$ has a local maximum and minimum at a point where ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$ that is $\mathrm{f}\left(\mathrm{x}\right)$ is concave up and concave down respectively and touches the $\mathrm{x}$ -axis.

From the graph provided in the question, it can be observed that $\mathrm{f}$ is concave up between $0\le \mathrm{x}\le 1$ and $3\le \mathrm{x}\le 5$ and is concave down between $1\le \mathrm{x}\le 3$ and $5\le \mathrm{x}\le 7$ .

Therefore,

The interval at which the function $\mathrm{g}$ is concave down is between $1\le \mathrm{x}\le 3$ and $5\le \mathrm{x}\le 7$ .

d.

To determine

To sketch the graph of $\mathrm{g}$

Explanation of Solution

Given information:

The function is $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}}$ and the graph of $\mathrm{f}\left(\mathrm{t}\right)$ is shown below:

  

Concept used:

The fundamental theorem of calculus, part 1 is defined by,

If $\mathrm{f}$ is continuous on $\left[\mathrm{a},\mathrm{b}\right]$ , then the function $\mathrm{g}$ defined by,

  $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{\mathrm{a}}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)}\mathrm{d}\mathrm{t}$ $\mathrm{a}\le \mathrm{x}\le \mathrm{b}$ is an antiderivative of $\mathrm{f}$ , that is ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ for $\mathrm{a}<\mathrm{x}<\mathrm{b}$ .

Graph:

From the graph of first derivative of $\mathrm{g}$ that is $\mathrm{f}\left(\mathrm{x}\right)$ , the following observation is made about the graph of $\mathrm{g}$ :

• The derivative ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ is zero at $\mathrm{x}=1,3,5,7,9$ , thus these are the turning point of the graph of the function $\mathrm{g}$ .
• The function $\mathrm{g}$ is increasing on the interval $\left(0,1\right),\left(3,5\right)\text{ and }\left(7,9\right)$ because its first derivative is positive at these interval.
• The function $\mathrm{g}$ is decreasing on the interval $\left(1,3\right)\text{ and }\left(5,7\right)$ because its first derivative is negative at these interval.

Hence, from the above observation,

The graph of $\mathrm{g}$ is obtained as: