Chapter 4.8, Problem 55E

(a)

To determine

To prove:

The function $H$ is constant.

Explanation of Solution

Given:

  $H\left(x\right)=G\left(x\right)-F\left(x\right)$

Concept used:

Antiderivative :- A function $F\left(x\right)$ is antiderivative of on an interval I if $F\text{'}\left(x\right)=f\left(x\right)$ for all $x$ in interval I

The entire family of antiderivative of a function by adding a constant to a known antiderivative

So, if $F\left(x\right)$ is the antiderivative of $f\left(x\right)$ , then the family of the antiderivatives would be $F\left(x\right)+c$

Calculation:

The mean-value theorem:- If $f$ is differential on the open interval $\left(a,b\right)$ and continuous on the closed interval $\left[a,b\right]$

Then there is atleast one number c in $\left(a,b\right)$ for which

  $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

  $\begin{array}{l}H\left(x_1\right)=G\left(x\right)-F\left(x\right)\\ H\left(x_2\right)=G\left(x\right)-F\left(x\right)\end{array}$

Here,

  $H\left(x_1\right)=H\left(x_2\right)=H$

So, $H$ satisfies all the conditions of the mean value theorem on $\left[x_1,x_2\right]$

Therefore $H$ is constant function.

(b)

To determine

To prove:

The function $H$ is constant.

Explanation of Solution

Given:

  $H\left(x\right)=G\left(x\right)-F\left(x\right)$

Concept used:

Antiderivative :- A function $F\left(x\right)$ is antiderivative of on an interval I if $F\text{'}\left(x\right)=f\left(x\right)$ for all $x$ in interval I

The entire family of antiderivative of a function by adding a constant to a known antiderivative

So, if $F\left(x\right)$ is the antiderivative of $f\left(x\right)$ , then the family of the antiderivatives would be $F\left(x\right)+c$

Calculation:

The mean-value theorem:- If $f$ is differential on the open interval $\left(a,b\right)$ and continuous on the closed interval $\left[a,b\right]$

Then there is atleast one number c in $\left(a,b\right)$ for which

  $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

  $\begin{array}{l}H\left(x_1\right)=G\left(x\right)-F\left(x\right)\\ H\left(x_2\right)=G\left(x\right)-F\left(x\right)\end{array}$

Here,

  $H\left(x_1\right)=H\left(x_2\right)=H$

So, $H$ satisfies all the conditions of the mean value theorem on $\left[x_1,x_2\right]$

Therefore $H$ is constant function.