Chapter 5, Problem 6RE

a. $$

To determine

Tofindthe interval on which the function $y=e^{x-1}-x$ is increasing.

Answer to Problem 6RE

The function is then increasing for $x\in \left(1,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=e^{x-1}-x$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=e^{x-1}-x$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=e^{x-1}-1$

  $y^{″}=e^{x-1}$

Set the derivative equal to zero,

  $\begin{array}{l}{y}^{\prime }=e^{x-1}-1=0\\ \Rightarrow x=1\end{array}$

Consider $x=2$ :

  $y^{\prime }\left(2\right)=e^{2-1}-1\approx 1.718>0$

Consider $x=0$ :

  $y^{\prime }\left(0\right)=e^{0-1}-1\approx -0.632<0$

Here, the function is increasing for $x\in \left(1,\infty \right)$ , $y^{\prime }>0$ .

Therefore, the function is increasing for $x\in \left(1,\infty \right)$ .

b. $$

To determine

To find the interval on which the function $y=e^{x-1}-x$ is decreasing.

Answer to Problem 6RE

The function is then decreasing for $x\in \left(-\infty ,1\right)$ .

Explanation of Solution

Given information:

The given function is $y=e^{x-1}-x$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=e^{x-1}-x$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=e^{x-1}-1$

  $y^{″}=e^{x-1}$

Set the derivative equal to zero,

  $\begin{array}{l}{y}^{\prime }=e^{x-1}-1=0\\ \Rightarrow x=1\end{array}$

Consider $x=2$ :

  $y^{\prime }\left(2\right)=e^{2-1}-1\approx 1.718>0$

Consider $x=0$ :

  $y^{\prime }\left(0\right)=e^{0-1}-1\approx -0.632<0$

Here, the function is decreasing for $x\in \left(-\infty ,1\right)$ , $y^{\prime }<0$ .

Therefore, the function is decreasing for $x>0.385$ .

c. $$

To determine

To find the interval on which the function $y=e^{x-1}-x$ is concave up.

Answer to Problem 6RE

The function is concave up for $x\in \left(-\infty ,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=e^{x-1}-x$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=e^{x-1}-x$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=e^{x-1}-1$

  $y^{″}=e^{x-1}$

Set the second derivative equal to zero,

  $y^{″}=e^{x-1}=no\text{ solution}$

Consider $x=2$ :

  $y^{″}=e^{2-1}=0$

Consider $x=-2$ :

  $y^{″}=e^{-2-1}=\text{no solution}$

Therefore, the function isconcave up for $x\in \left(-\infty ,\infty \right)$ .

d. $$

To determine

To find the interval on which the function $y=e^{x-1}-x$ is concave down.

Answer to Problem 6RE

The function is neverconcave down.

Explanation of Solution

Given information:

The given function is $y=e^{x-1}-x$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=e^{x-1}-x$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=e^{x-1}-1$

  $y^{″}=e^{x-1}$

Set the second derivative equal to zero,

  $y^{″}=e^{x-1}=no\text{ solution}$

Consider $x=2$ :

  $y^{″}=e^{2-1}=0$

Consider $x=-2$ :

  $y^{″}=e^{-2-1}=\text{no solution}$

  $y^{″}>0$ for any $x$ , the function is never concave down.

Therefore, the function is never concave down.

e. $$

To determine

To find the interval on which the function $y=e^{x-1}-x$ has local extreme values.

Answer to Problem 6RE

The function haslocal maximum at $x=1$ .

Explanation of Solution

Given information:

The given function is $y=e^{x-1}-x$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=e^{x-1}-x$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=e^{x-1}-1$

  $y^{″}=e^{x-1}$

Set the derivative equal to zero,

  $\begin{array}{l}{y}^{\prime }=e^{x-1}-1=0\\ \Rightarrow x=1\end{array}$

At critical point $x=1$ , $y^{\prime }$ changes sign from negative to positive. So $y$ has a local minimum at $x=1$ , where

  $\begin{array}{l}y=e^{x-1}-x\\ y\left(1\right)=e^{1-1}-1=0\end{array}$

Therefore, the function has local minimum at $x=1$ .

f. $$

To determine

To find the interval on which the function $y=e^{x-1}-x$ has inflection points.

Answer to Problem 6RE

The function has noinflection point.

Explanation of Solution

Given information:

The given function is $y=e^{x-1}-x$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=e^{x-1}-x$ ,

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $y^{\prime }=e^{x-1}-1$

  $y^{″}=e^{x-1}$

Set the second derivative equal to zero,

  $\begin{array}{l}{y}^{″}=e^{x-1}=\text{no solution}\\ \Rightarrow x=1\end{array}$

Since $y^{″}>0$ for any $x$ , the function has no inflection points.

Therefore, the function has no inflection point.