Chapter 5, Problem 48RE

a.

To determine

To identify: Whether the statement is true or not “ g is the differentiable function of x ”. It is given that $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}\text{ and }f\left(1\right)=0$ , also $f\left(x\right)$ has positive derivative.

Answer to Problem 48RE

The given statement is True.

Explanation of Solution

Given information:

The given integral is $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}$ .

Consider the given Integral.

It is given that $f\left(1\right)=0$ it means $f\text{'}\left(1\right)>0$ for all real number.

  $f\left(x\right)$ is increasing for all real number. So $f\left(x\right)<0$ for $x<1$ .

Now differentiate the given function.

  $\begin{array}{c}g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}\\ g\text{'}\left(x\right)=f\left(x\right)\frac{d}{dx}\left(x\right)-f\left(0\right)\frac{d}{dx}\left(0\right)\\ g\text{'}\left(x\right)=f\left(x\right)\end{array}$

It is given that $f\left(x\right)$ has positive derivative so it means $f\left(x\right)$ is continuous because if a function is differentiable then it must be continuous.

Yes, $g\left(x\right)$ is differentiable.

b.

To determine

To identify: Whether the statement is true or not “ g is continuous function of x ”. It is given that $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}\text{ and }f\left(1\right)=0$ , also $f\left(x\right)$ has positive derivative.

Answer to Problem 48RE

The given statement is True.

Explanation of Solution

Given information:

The given integral is $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}$ .

Consider the given Integral.

From part (a) it is known that $g\text{'}\left(x\right)=f\left(x\right)$ . Also $g\left(x\right)$ is differentiable so the function must be continuous because if a function is differentiable then it must be continuous.

Yes, g is continuous function of x .

c.

To determine

To identify: Whether the statement is true or not “The graph of g has a horizontal tangent line at $x=1$ ”. It is given that $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}\text{ and }f\left(1\right)=0$ , also $f\left(x\right)$ has positive derivative.

Answer to Problem 48RE

The given statement is True.

Explanation of Solution

Given information:

The given integral is $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}$ .

Consider the given Integral.

Horizontal tangent line means $g\text{'}\left(x\right)=0$ .

From part (a) it is known that $g\text{'}\left(x\right)=f\left(x\right)$ .

Also, it is given that $f\left(1\right)=0$ it means $g\text{'}\left(1\right)=0$ .

Thus, the graph of g has a horizontal tangent line at $x=1$ .

Yes, the graph of g has a horizontal tangent line at $x=1$ .

d.

To determine

To identify: Whether the statement is true or not “ g has a local maximum at $x=1$ ”. It is given that $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}\text{ and }f\left(1\right)=0$ , also $f\left(x\right)$ has positive derivative.

Answer to Problem 48RE

The given statement is False.

Explanation of Solution

Given information:

The given integral is $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}$ .

Consider the given Integral.

From part (a) it is known that $g\text{'}\left(x\right)=f\left(x\right)$ .

Thus,

  $\begin{array}{c}g\text{'}\text{'}\left(x\right)=f\text{'}\left(x\right)\\ g\text{'}\text{'}\left(1\right)=f\text{'}\left(1\right)>0\end{array}$

Thus, the given statement is False.

e.

To determine

To identify: Whether the statement is true or not “ g has a local minimum at $x=1$ ”. It is given that $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}\text{ and }f\left(1\right)=0$ , also $f\left(x\right)$ has positive derivative.

Answer to Problem 48RE

The given statement is True.

Explanation of Solution

Given information:

The given integral is $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}$ .

Consider the given Integral.

From part (a) it is known that $g\text{'}\left(x\right)=f\left(x\right)$ .

Thus,

  $\begin{array}{c}g\text{'}\text{'}\left(x\right)=f\text{'}\left(x\right)\\ g\text{'}\text{'}\left(1\right)=f\text{'}\left(1\right)>0\end{array}$

At $x=1,g\left(x\right)$ has local Maximum.

The given statement is True.

f.

To determine

To identify: Whether the statement is true or not “The graph of g has an inflation point at $x=1$ ”. It is given that $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}\text{ and }f\left(1\right)=0$ , also $f\left(x\right)$ has positive derivative.

Answer to Problem 48RE

The given statement is False.

Explanation of Solution

Given information:

The given integral is $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}$ .

Consider the given Integral.

From part (a) it is known that $g\text{'}\left(x\right)=f\left(x\right)$ .

Since, at $x=1,g\left(x\right)$ has local Maximum.

The given statement is False.

g.

To determine

To identify: Whether the statement is true or not “The graph of $\frac{dg}{dx}$ crosses the x -axis at $x=1$ ”. It is given that $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}\text{ and }f\left(1\right)=0$ , also $f\left(x\right)$ has positive derivative.

Answer to Problem 48RE

The given statement is True.

Explanation of Solution

Given information:

The given integral is $g\left(x\right)={\displaystyle {\int }_0^{x}f\left(t\right)dt}$ .

Consider the given Integral.

From part (a) it is known that $g\text{'}\left(1\right)=f\left(1\right)=0$ and $g\text{'}\left(x\right)=f\left(x\right)$ is an increasing function so the graph of $\frac{dg}{dx}$ crosses the x -axis at $x=1$ .

The given statement is True.