Chapter 6.4, Problem 75E

(a)

To determine

Whether h is a twice-differentiable function of x.

Answer to Problem 75E

The statement is true.

Explanation of Solution

Given information:

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

Where,

f has a negative derivative for all values of x and that $f\left(1\right)=0$ .

According to FTC (Fundamental Theorem of Calculus),

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

Then

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

Taking derivative again

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

According to the statement,

f has negative derivative for all values of x.

That means

  $f\text{'}\left(x\right)$ is defined for all x.

Then

  $h\text{'}\text{'}\left(x\right)$ must also be defined for all x.

Therefore,

h is a twice differentiable function of x.

(b)

To determine

Whether h and $dh/dx$ are both continuous.

Answer to Problem 75E

The statement is true.

Explanation of Solution

Given information:

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

Where,

f has a negative derivative for all values of x and that $f\left(1\right)=0$ .

According to FTC (Fundamental Theorem of Calculus),

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

Then

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

Taking derivative again

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

Thus,

We have

Both $h\left(x\right)$ and $h\text{'}\left(x\right)$ are differentiable.

Therefore,

They both are continuous.

(c)

To determine

Whether the graph of h has a horizontal tangent at $x=1$ .

Answer to Problem 75E

The statement is true.

Explanation of Solution

Given information:

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

Where,

f has a negative derivative for all values of x and that $f\left(1\right)=0$ .

If $h\text{'}\left(1\right)=0$ ,

h has a horizontal tangent at $x=1$ .

Since

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

Then

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

According to the statement,

  $f\left(1\right)=0$

Therefore,

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

(d)

To determine

Whether h has local maximum at $x=1$ .

Answer to Problem 75E

The statement is true by the Second Derivative Test.

Explanation of Solution

Given information:

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

Where,

f has a negative derivative for all values of x and that $f\left(1\right)=0$ .

If $h\text{'}\left(1\right)=0$

And

  $h\text{'}\text{'}\left(1\right)<0$

Then

h has a local maximum at $x=1$ .

According to Part (c) result,

We have

  $h\text{'}\left(1\right)=0$

Since

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

Then

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

According to the statement,

f has negative derivative for all values of x.

That means

  $h\text{'}\text{'}\left(x\right)=f\text{'}\left(x\right)<0$ for all values of x.

Therefore,

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

(e)

To determine

Whether h has local minimum at $x=1$ .

Answer to Problem 75E

The statement is false by the Second Derivative Test.

Explanation of Solution

Given information:

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

Where,

f has a negative derivative for all values of x and that $f\left(1\right)=0$ .

If $h\text{'}\left(1\right)=0$

And

  $h\text{'}\text{'}\left(1\right)>0$

Then

h has a local minimum at $x=1$ .

According to Part (a) result,

We have

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

And

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

Then

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

And

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

According to statement,

  $f\left(1\right)=0$

Thus,

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

We also have

  $h\text{'}\left(x\right)<0$

Thus,

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

Therefore,

h does not have a local minimum at $x=1$ .

(f)

To determine

Whether the graph of h has an inflection point at $x=1$ .

Answer to Problem 75E

The statement is false.

Explanation of Solution

Given information:

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

Where,

f has a negative derivative for all values of x and that $f\left(1\right)=0$ .

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

Then

h has an inflection point at $x=1$ .

According to Part (a) result,

We have

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

And

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

Then

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

And

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

We also have

  $h\text{'}\left(x\right)<0$

Thus,

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

Therefore,

h does not have an inflection point at $x=1$ .

(g)

To determine

Whether the graph $dh/dx$ crosses at x-axis at $x=1$ .

Answer to Problem 75E

The statement is true.

Explanation of Solution

Given information:

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

Where,

f has a negative derivative for all values of x and that $f\left(1\right)=0$ .

We have

  $\frac{dh}{dx}=h\text{'}\left(x\right)=f\left(0\right)$

According to the statement,

f is a decreasing function.

That includes

Point (1, 0) because $f\left(1\right)=0$ .

Then

  $h\text{'}\left(x\right)$ also includes the point (1, 0).

Therefore,

  $h\text{'}\left(x\right)$ crosses the x-axis at $x=1$ .