Chapter 6, Problem 54RE

(a)

To determine

Evaluate $g\left(x\right)$ at $x=1$ .

Answer to Problem 54RE

At x = 1:

  $g\left(1\right)=0$

Explanation of Solution

Given information:

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

The graph of a function f :

  

Calculation:

We have

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

At $x=1$ :

  $g\left(1\right)={\displaystyle {\int }_1^{1}f\left(t\right)dt}$

By evaluating the integral:

  $g\left(1\right)={\displaystyle {\int }_1^{1}f\left(t\right)dt}=0$

Therefore,

  $g\left(1\right)=0$

(b)

To determine

Evaluate $g\left(x\right)$ at $x=3$ .

Answer to Problem 54RE

At $x=3$ :

  $g\left(3\right)=-1$

Explanation of Solution

Given information:

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

The graph of a function f :

  

Calculation:

Since

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

Then

  $g\left(3\right)={\displaystyle {\int }_1^{3}f\left(t\right)dt}$

The area of the graph between 1 and 3 is a triangle below the x − axis.

Thus,

It has a negative area.

That includes

Base of 2 and Height of 1.

Then

The area:

  $g\left(3\right)={\displaystyle {\int }_1^{3}f\left(t\right)dt}=-\frac{1}{2}\cdot 2\cdot 1=-1$

(c)

To determine

Evaluate $g\left(x\right)$ at $x=-1$ .

Answer to Problem 54RE

At $x=-1$ :

  $g\left(-1\right)=-\pi$

Explanation of Solution

Given information:

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

The graph of a function f :

  

Calculation:

We have

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

Substitute $x=-1$ :

  $g\left(-1\right)={\displaystyle {\int }_1^{-1}f\left(t\right)dt}$

To switch the bounds, use the Order of Integration rule:

  $g\left(-1\right)={\displaystyle {\int }_{-1}^{1}f\left(t\right)dt}$

  $\left[{\displaystyle {\int }_a^{b}f\left(x\right)dx}=-{\displaystyle {\int }_b^{a}f\left(x\right)dx}\right]$

The region from $x=-1$ and $x=1$ is a quarter circle with a radius of 2.

Then

Use the formula for the area of the quarter circle:

  $A=\frac{1}{4}\pi r^2$

Therefore,

  $\begin{array}{l}g\left(-1\right)=-{\displaystyle {\int }_{-1}^{1}f\left(t\right)dt}\\ =-\frac{1}{4}\pi {\left(2\right)}^2\\ =-\pi \end{array}$

(d)

To determine

All values of x on the open interval [-3, 4] at which g has a relative maximum.

Answer to Problem 54RE

  $g\left(x\right)$ has a relative maximum at $x=1$ .

Explanation of Solution

Given information:

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

The graph of a function f :

  

Calculation:

To know where $g\left(x\right)$ has a relative maximum, we are required to know where g’ switches from positive to negative.

Since

  $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}$

According to FTC:

  $g\text{'}\left(x\right)=\frac{d}{dx}{\displaystyle {\int }_0^{x}f\left(t\right)dt}=f\left(x\right)$

Then

  $f\left(x\right)>0$ for all $x\in \left(-3,1\right)$

And

  $f\left(x\right)<0$ for all $x\in \left(1,3\right)$

Therefore,

  $g\left(x\right)$ has a relative maximum at $x=1$ .

(e)

To determine

Equation for the line tangent to the graph of g at $x=-1$

Answer to Problem 54RE

Linear function:

  $y+\pi =2\left(x+1\right)$

Explanation of Solution

Given information:

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

The graph of a function f :

  

Calculation:

The slope of linear function:

  $m=g\text{'}\left(-1\right)=f\left(-1\right)=2$

From Part (c) result,

We have

  $g\left(-1\right)=-\pi$

Then

The Linear function:

  $y-\left(-\pi \right)=2\left(x-\left(-1\right)\right)=2\left(x+1\right)$

Rewrite the above equation:

Therefore,

  $y+\pi =2\left(x+1\right)$

(f)

To determine

x − coordinate of each point of inflection of the graph of g at $x=-1$

Answer to Problem 54RE

x − coordinates of the points of inflection:

  $x=-1\text{and}x=2$

Explanation of Solution

Given information:

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

The graph of a function f :

  

Calculation:

The points of inflection of g occur at x − values.

Such that

  $g"=0$ or is undefined and the sign of g” changes.

Since

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

Then

  $g\text{'}\left(x\right)=f\left(x\right)$

That further becomes

  $g\text{'}\text{'}\left(x\right)=f\text{'}\left(x\right)$

Therefore,

The values of x that make $g\text{'}\text{'}=0$ or be undefined and make g” change sign are the values of x that make $f\text{'}\left(x\right)=0$ or undefined and change the sign of f’ .

From the graph of f ,

The slope of f is 0 at $x=-1$ .

Thus,

  $g"\left(-1\right)=f\text{'}\left(-1\right)=0$

And

The slope of f is undefined at $x=1$ (which is a cusp) and $x=2$ (which is a corner) making g” undefined.

Thus,

f’ switches from positive to negative at $x=-1$ , it is a point of inflection.

Then

f’ is negative to the left and right of $x=1$ , it is not a point of inflection.

Then

f’ switches from negative to positive at $x=2$ , it is a point of inflection.

Therefore,

The x − coordinates of the points of inflection are $x=-1$ and $x=2$ .

(g)

To determine

Range of g

Answer to Problem 54RE

The range is [ $-2\pi$ , 0].

Explanation of Solution

Given information:

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

The graph of a function f :

  

Calculation:

In order to obtain the range of g ,

We are required to find the absolute minimum absolute maximum of g .

To obtain absolute extremes,

We are required to find the critical values and evaluate those critical values and endpoints in the original function to know which gives the largest value and which gives the least.

Since

  $g\left(x\right)={\displaystyle {\int }_1^{x}f\left(t\right)dt}$

Then

  $g\text{'}\left(x\right)=\frac{d}{dx}{\displaystyle {\int }_1^{x}f\left(t\right)dt}=f\left(x\right)$

From the graph,

  $g\text{'}\left(x\right)=f\left(x\right)=0$

At

  $x=1\text{and}x=3$

And

At the endpoints:

  $x=-3\text{and}x=4$

At $x=-3$ :

  $g\left(-3\right)={\displaystyle {\int }_1^{-3}f\left(t\right)dt}=-{\displaystyle {\int }_{-3}^{1}f\left(t\right)dt}=-\frac{1}{2}\pi {\left(2\right)}^2=-2\pi$

At $x=1$ :

  $g\left(1\right)={\displaystyle {\int }_1^{1}f\left(t\right)dt}=0$

At $x=3$ :

  $g\left(3\right)={\displaystyle {\int }_1^{3}f\left(t\right)dt}=\frac{1}{2}\left(2\right)\left(-1\right)=-1$

At $x=4$ :

  $g\left(4\right)={\displaystyle {\int }_1^{4}f\left(t\right)dt}=\frac{1}{2}\left(2\right)\left(-1\right)+\frac{1}{2}\left(1\right)\left(1\right)=-1+\frac{1}{2}=-\frac{1}{2}$

Therefore,

Absolute minimum is $-2\pi$ .

And

Absolute maximum is 0.

Then

The range is [ $-2\pi$ , 0].