Chapter 5.3, Problem 12E

a.

To determine

To determine the intervals on which the graph of the function is concave up.

Answer to Problem 12E

Concave up at $\left[0,2x\right]$

Explanation of Solution

Given Information:

Given equation is,

  $y=e^x,0\le x\le 2\pi$

Calculation:

The graph of a differentiable function $y=f\left(x\right)$ is concave up on an open interval $I$ if $y\text{'}$ is increasing on $I$ .

Where if a function $y=f\left(x\right)$ has a second derivative then it can conclude that $y\text{'}$ increases if $y\text{'}\text{'}>0$ and decreases if $y\text{'}\text{'}<0$ .

  $\begin{array}{l}y=e^x\\ y^{\prime }=e^x\\ y^{″}=e^x\\ x=0,e^0=1\\ x=1,e^1\approx 2.71828183\end{array}$

The $y^{\prime }$ and $y^{″}$ is positive,

So, $y$ is always increasing.

Hence,

Concave up at $\left[0,2x\right]$

b.

To determine

To determine the intervals on which the graph of the function is concave down.

Answer to Problem 12E

Concave up at $\left[0,2x\right]$

Explanation of Solution

Given Information:

The given equation is,

  $y=e^x,0\le x\le 2\pi$

Calculation:

The graph of a differentiable function $y=f\left(x\right)$ is concave down on an open interval $I$ if $y\text{'}$ is increasing on $I$ .

Where if a function $y=f\left(x\right)$ has a second derivative then it can conclude that $y\text{'}$ increases if $y\text{'}\text{'}>0$ and decreases if $y\text{'}\text{'}<0$ .

  $\begin{array}{l}y=e^x\\ y^{\prime }=e^x\\ y^{″}=e^x\\ x=0,e^0=1\\ x=1,e^1\approx 2.71828183\end{array}$

The $y^{\prime }$ And $y^{″}$ is positive, so $y$ is always increasing and never concave down

Hence,

Not concave down.