Chapter 2.8, Problem 22E

To determine

To graph:The function $F\left(z\right)=\frac{z^3}{e^z}$ for $0\le z\le 5$ with help of first and second derivative of the function and identify the regions in which function is increasing and concave up.

Explanation of Solution

Given information:

The function $F\left(z\right)=\frac{z^3}{e^z}$ and domain $0\le z\le 5$ .

Graph:

Consider the provided function $F\left(z\right)=\frac{z^3}{e^z}$ .

Evaluate the first derivative of the function.

Apply the quotient rule of differentiation, $\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{g\left(x\right)\frac{d}{dx}f\left(x\right)-f\left(x\right)\frac{d}{dx}g\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ .

  $\begin{array}{c}\frac{d}{dz}F\left(z\right)=\frac{d}{dz}\left(\frac{z^3}{e^z}\right)\\ =\frac{e^z\frac{d}{dz}\left(z^3\right)-z^3\frac{d}{dz}\left(e^z\right)}{e^{2z}}\\ =\frac{3z^2{e}^z-e^z{z}^3}{e^{2z}}\\ =\frac{z^2{e}^z\left(3-z\right)}{e^{2z}}\end{array}$

Simplify it further as,

  $\frac{d}{dz}F\left(z\right)=\frac{z^2\left(3-z\right)}{e^z}$

Recall if first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

The first derivative is positive if,

  $\begin{array}{c}\frac{z^2\left(3-z\right)}{e^z}>0\\ 3-z>0\\ 3>z\\ z<3\end{array}$

The function is increasing if $z<3$ .

The first derivative is negative if,

  $\begin{array}{c}\frac{z^2\left(3-z\right)}{e^z}<0\\ 3-z<0\\ 3<z\end{array}$

The function is decreasing if $z>3$ .

Therefore, the function is increasing when $z<3$ and decreasing when $z>3$ .

Evaluate the second derivative of the function, differentiate first derivative again with respect to x .

Apply the quotient rule of differentiation, $\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{g\left(x\right)\frac{d}{dx}f\left(x\right)-f\left(x\right)\frac{d}{dx}g\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ .

  $\begin{array}{c}\frac{d^2}{d{z}^2}F\left(z\right)=\frac{d}{dz}\left(\frac{z^2\left(3-z\right)}{e^z}\right)\\ =\frac{e^z\frac{d}{dz}\left(z^2\left(3-z\right)\right)-z^2\left(3-z\right)\frac{d}{dz}\left(e^z\right)}{e^{2z}}\\ =\frac{e^z\frac{d}{dz}\left(z^2\left(3-z\right)\right)-e^z{z}^2\left(3-z\right)}{e^{2z}}\end{array}$

Apply the product rule, $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=f\left(x\right)\frac{d}{dx}g\left(x\right)+g\left(x\right)\frac{d}{dx}f\left(x\right)$

  $\begin{array}{c}\frac{d^2}{d{z}^2}F\left(z\right)=\frac{e^z\frac{d}{dz}\left(z^2\left(3-z\right)\right)-e^z{z}^2\left(3-z\right)}{e^{2z}}\\ =\frac{e^z\left[-z^2+\left(3-z\right)2z\right]-3z^2{e}^z+z^3{e}^z}{e^{2z}}\\ =\frac{-3z^2{e}^z+6z{e}^z-3z^2{e}^z+z^3{e}^z}{e^{2z}}\\ =\frac{-6z^2{e}^z+6z{e}^z+z^3{e}^z}{e^{2z}}\end{array}$

Simplify it further as,

  $\begin{array}{c}F\text{'}\text{'}\left(z\right)=\frac{-6z^2{e}^z+6z{e}^z+z^3{e}^z}{e^{2z}}\\ =\frac{z{e}^z\left(z^2-6z+6\right)}{e^{2z}}\\ =\frac{z\left(z^2-6z+6\right)}{e^z}\end{array}$

If second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is concave up on interval $\left[c,d\right]$ .

If second derivative of the function is less than zero that is $g\text{'}\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is concave down on interval $\left[c,d\right]$ .

The second derivative is always positive so the function is always concave up.

Therefore, the graph of the function $F\left(z\right)=\frac{z^3}{e^z}$ for $0\le z\le 5$ is provided below,

  

Interpretation:

Observe that first derivative of the function is $F\text{'}\left(z\right)=\frac{z^2\left(3-z\right)}{e^z}$ which is positive if, x is less than 3 and negative if x is greater than 3.

Therefore, the function $F\left(z\right)=\frac{z^3}{e^z}$ is increasing when $z<3$ and decreasing when $z>3$ .

Next observe that second derivative of the function is $F\text{'}\text{'}\left(z\right)=\frac{z\left(z^2-6z+6\right)}{e^z}$ which is positive, so function $F\left(z\right)=\frac{z^3}{e^z}$ is concave up always.

Thus, the function, $F\left(z\right)=\frac{z^3}{e^z}$ is increasing when $z<3$ and concave up always.