Chapter 6, Problem 46RE

(a)

To determine

Whether g is differentiable function of x .

Answer to Problem 46RE

The statement is True.

Explanation of Solution

Given information:

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

  $f\left(x\right)$ has positive derivative for all values of x .

Such that

  $f\left(1\right)=0$

g is differentiable function if g’ is continuous.

Since

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

Then

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

According the statement,

  $f\left(x\right)$ is differentiable for all x .

Thus,

  $f\left(x\right)$ is continuous for all x .

Since

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

Then

  $g\text{'}\left(x\right)$ must also be continuous.

Therefore,

g is a differentiable function of x .

(b)

To determine

Whether g is a continuous function of x

Answer to Problem 46RE

The statement is True.

Explanation of Solution

Given information:

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

  $f\left(x\right)$ has positive derivative for all values of x .

Such that

  $f\left(1\right)=0$

Since

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

Then

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

According the statement,

  $f\left(x\right)$ is differentiable for all x .

Thus,

  $f\left(x\right)$ is continuous for all x .

Since

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

Then

  $g\text{'}\left(x\right)$ must also be continuous.

We have already discussed in Part (a),

g is differentiable function if g’ is continuous.

Therefore,

g is a continuous function of x .

(c)

To determine

Whether the graph of g has a horizontal tangent line at $x=1$

Answer to Problem 46RE

The statement is True.

Explanation of Solution

Given information:

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

  $f\left(x\right)$ has positive derivative for all values of x .

Such that

  $f\left(1\right)=0$

Since

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

Then

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

According the statement,

  $f\left(1\right)=0$

Then

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

Therefore,

The graph of g has a horizontal tangent line at $x=1$ .

(d)

To determine

Whether g has a local maximum at $x=1$

Answer to Problem 46RE

The statement is False.

Explanation of Solution

Given information:

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

  $f\left(x\right)$ has positive derivative for all values of x .

Such that

  $f\left(1\right)=0$

If $g\left(x\right)$ has a local maximum at $x=1$ ,

Then

By the Second Derivative Test:

  $g\text{'}\left(1\right)=0$

And

  $g\text{'}\text{'}\left(1\right)<0$

Since

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

Then

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

According the statement,

  $f\left(1\right)=0$

Then

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

We also have

  $f\text{'}\left(x\right)>0$ for all x .

Such that

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

Therefore,

  $g\left(x\right)$ does not have a local maximum at $x=1$ .

(e)

To determine

Whether g has a local minimum at $x=1$

Answer to Problem 46RE

The statement is True.

Explanation of Solution

Given information:

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

  $f\left(x\right)$ has positive derivative for all values of x .

Such that

  $f\left(1\right)=0$

If $g\left(x\right)$ has a local minimum at $x=1$ ,

Then

By the Second Derivative Test:

  $g\text{'}\left(1\right)=0$

And

  $g\text{'}\text{'}\left(1\right)>0$

Since

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

Then

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

According the statement,

  $f\left(1\right)=0$

Then

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

We also have

  $f\text{'}\left(x\right)>0$ for all x .

Such that

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

Thus,

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

Therefore,

  $g\left(x\right)$ does have a local minimum at $x=1$ .

(f)

To determine

Whether the graph of g has an infection point at $x=1$

Answer to Problem 46RE

The statement is False.

Explanation of Solution

Given information:

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

  $f\left(x\right)$ has positive derivative for all values of x .

Such that

  $f\left(1\right)=0$

If $g\left(x\right)$ has an inflection point at $x=1$ ,

Then

  $g"\left(1\right)=0$

However,

  $g"\left(x\right)=f\text{'}\left(x\right)>0$ for all values of x .

Such that

  $g"\left(1\right)>0$

Thus,

  $g\left(x\right)$ does not have an inflection point at $x=1$ .

(g)

To determine

Whether the graph of $dg/dx$ crosses the x − axis at $x=1$

Answer to Problem 46RE

The statement is True.

Explanation of Solution

Given information:

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

  $f\left(x\right)$ has positive derivative for all values of x .

Such that

  $f\left(1\right)=0$

Since

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

And

  $f\left(1\right)=0$

Then

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

  $\frac{dg}{dx}$ must contain the point (1,0).

Since

  $f\text{'}\left(x\right)>0$ for all x .

Then

  $f\left(x\right)$ is a strictly increasing function.

Such that

Instead of the alternative of bouncing − off of the x − axis at (1, 0),

  $f\left(x\right)$ crosses the x − axis at (1, 0).

Therefore,

  $\frac{dg}{dx}$ crosses the x − axis at (1, 0).