Chapter 4, Problem 1RE

(a) If $x_1<x_2,$ what relationship must hold between $f\left(x_1\right)$ and $f\left(x_2\right)$ if $f$ in increasing on an interval containing $x_1$ and $x_2$ ? Decreasing? Constant?

(b) What condition on $f^{\prime }$ ensures that $f$ is increasing on an interval $\left[a,b\right]$ ? Decreasing? Constant?

To determine

The relationship must hold between $f\left(x_1\right)$ and $f\left(x_2\right)$ for $x_1<x_2$ if $f$ is increasing on an interval containing $x_1$ and $x_2$ , decreasing constant.

Answer to Problem 1RE

The function $f$ is increasing on the interval, if $f\left(x_1\right)<f\left(x_2\right)$ whenever $x_1<x_2$ .

The function $f$ is decreasing on the interval, if $f\left(x_1\right)>f\left(x_2\right)$ whenever $x_1<x_2$ .

The function $f$ is constant on the interval, if $f\left(x_1\right)=f\left(x_2\right)$ whenever $x_1$ and $x_2$ .

Explanation of Solution

The function $f$ is increasing on the interval, if $f\left(x_1\right)<f\left(x_2\right)$ whenever $x_1<x_2$ .

For example:

$f\left(x\right)=e^x$

Now, draw the graph of $e^x$ .

It can be observed from the graph that the function value is increasing, if $x$ is increasing.

The function $f$ is decreasing on the interval, if $f\left(x_1\right)>f\left(x_2\right)$ whenever $x_1<x_2$ .

For example:

$f\left(x\right)=e^{-x}$

Now, draw the graph of $e^{-x}$ .

It can be observed from the graph that the function value is decreasing if $x$ is increasing.

The function $f$ is constant on the interval, if $f\left(x_1\right)=f\left(x_2\right)$ whenever $x_1$ and $x_2$ .

$f\left(x\right)=5$

Now, draw the graph of $f\left(x\right)=5$ .

It can be observed from the graph that the function value is constant if $x$ is increasing.

To determine

The condition on $f^{\text{'}}$ ensures that $f$ is increasing ,decreasing ,constant on an interval $\left[a,b\right]$ .

Answer to Problem 1RE

If $f^{\text{'}}>0$ on the interval, then $f$ is increasing.

If $f^{\text{'}}<0$ , then $f$ is decreasing.

If $f^{\text{'}}=0$ on the interval, then $f$ is constant.

Explanation of Solution

Consider the given statement.

$f$ is increasing ,decreasing ,constant on an interval $\left[a,b\right]$ .

If $f^{\text{'}}>0$ on the interval, then $f$ is increasing.

Consider the example $f\left(x\right)=x^2$ on the interval $\left[0,\infty \right]$ .

$f^{\prime }\left(x\right)=2x$

$f^{\prime }\left(x\right)>0$ for the interval $\left[0,\infty \right]$

So, $f\left(x\right)=x^2$ is increasing on the interval $\left[0,\infty \right]$ .

If $f^{\text{'}}<0$ , then $f$ is decreasing.

Consider the example $f\left(x\right)=x^2$ on the interval $\left[-\infty ,0\right]$ .

$f^{\prime }\left(x\right)=2x$

$f^{\prime }\left(x\right)<0$ for the interval $\left[-\infty ,0\right]$

So, $f\left(x\right)=x^2$ is decreasing on the interval $\left[-\infty ,0\right]$ .

If $f^{\text{'}}=0$ on the interval, then $f$ is constant.

Consider the example $f\left(x\right)=5$ on the interval $\left[-\infty ,\infty \right]$ .

$f^{\prime }\left(x\right)=0$

So, $f\left(x\right)=5$ is constant on the interval $\left[-\infty ,\infty \right]$ .