Chapter 6.R, Problem 82E

To determine

To Draw:

Graph of the function $y=e^{2x-x^2}$.

Solution:

The graph is drawn below.

Given: $y=e^{2x-x^2}$

Explanation:

Consider, the function

$y=e^{2x-x^2}$

(1) Domain of the function y:

Domain of $e^x\text{ is }ℝ$. Therefore,

Domain of $y=e^{2x-x^2}\text{ is also }ℝ$.

(2) Intercepts:

For y-intercept, put x = 0 in the function y:

 $\begin{array}{c}y(0)=e^{2\times 0-0^2}\\ =e^0\\ =1\end{array}$

Therefore, (0,1) is y-intercept. Since range of exponential function is $\left(0,\infty \right)$, therefore, y has no x-intercept.

(3)

$\begin{array}{c}y(-x)=e^{-2x-x^2}\\ \ne -y(x)\\ \ne y(x)\end{array}$

This implies that y is neither even nor odd function.

(4) Asymptotes:

Since the function y is finite for all values, therefore, it has no vertical asymptote.

Horizontal asymptotes:

Consider

$\begin{array}{c}\underset{x\to \infty }{\lim }y(x)=\underset{x\to \infty }{\lim }{e}^{2x-x^2}\\ =e^{-\infty }\\ =0\end{array}$

Thus, y =0 is horizontal asymptotes.

(5) Increasing or decreasing:

Consider

$y=e^{2x-x^2}$

Differentiate it with respect to x, we get

$\begin{array}{c}y\text{'}=\frac{d}{dx}{e}^{2x-x^2}\\ =\frac{d}{dx}{e}^u\left(\text{Put }u=2x-x^2\right)\\ =\frac{d}{du}{e}^u\frac{du}{dx}\text{ (By chain rule)}\\ \text{=}{e}^u\frac{d}{dx}(u)\\ =e^{2x-x^2}\frac{d}{dx}\left(2x-x^2\right)\left(\text{Replace the value of }u\right)\\ =e^{2x-x^2}\times \left(2-2x\right)\\ =\left(2-2x\right)e^{2x-x^2}\end{array}$

Put $y\text{'}(x)=0$. This implies that

$\begin{array}{l}\left(2-2x\right)e^{2x-x^2}=0\\ \text{So, }\\ 2-2x=0\text{ as exponential function is always positive}\text{.}\\ \text{This gives that}\\ 1-x=0\\ \text{Thus, }x=1\end{array}$

Consider

Interval	Point in the interval	Value of $\left(2-2x\right)e^{2x-x^2}$	Sign of $y\text{'}$	Increasing or decreasing
$(-\infty ,1)$	$0$	$e^{2\times 0-0^2}\left(2-2\times 0\right)=2$	$\text{+ve}$	Increasing
$(1,\infty )$	$2$	$\begin{array}{l}{e}^{2\times 2-2^2}(2-2\times 2)=e^0\times (-2)\\ \text{ =}-\text{2}\end{array}$	$-\text{ve}$	Decreasing

This implies that $y$ is increasing in the interval $(-\infty ,1)$ and decreasing in $(1,\infty )$.

(6) Local maxima or minima:

Consider

$y\text{'}=\left(2-2x\right)e^{2x-x^2}$

Differentiate it with respect to x, we get

$\begin{array}{c}y\text{'}\text{'}=\frac{d}{dx}\left(\left(2-2x\right)e^{2x-x^2}\right)\\ =\frac{d}{dx}\left(2-2x\right)e^{2x-x^2}\text{+}(2-2x)\frac{d}{dx}{e}^{2x-x^2}\end{array}$

  $\begin{array}{c}\text{=}(0-2)e^{2x-x^2}+(2-2x)e^{2x-x^2}\frac{d}{dx}\left(2x-x^2\right)\\ =e^{2x-x^2}\left[-2+(2-2x)(2-2x)\right]\\ =2e^{2x-x^2}\left[-1+2{(1-x)}^2\right]\end{array}$

At $x=e^{-1}$

$\begin{array}{l}y\text{'}\text{'}\left(1\right)=2e^1\left[-1+2{(1-1)}^2\right]\\ \text{ =}-2e<0\end{array}$

Therefore, $x=1$ is point of maxima.

(7) Inflection point:

Put $y\text{'}\text{'}(x)=0$, we get

$\begin{array}{l}0=y\text{'}\text{'}(x)\\ =2e^{2x-x^2}\left[-1+2{(1-x)}^2\right]\\ \text{This implies that}\\ -1+2{(}^1=0\\ \Rightarrow {(}^1=\frac{1}{2}\\ \Rightarrow 1-x=\pm \frac{1}{\sqrt{2}}\\ \Rightarrow x=1\pm \frac{1}{\sqrt{2}}\approx 1.71\text{ or 0}\text{.29}\end{array}$

Now,

$\begin{array}{l}y\text{'}\text{'}\left(0\right)=2e^0\left[-1+2{(1-0)}^2\right]\\ \text{ =}2>0\end{array}$

$\begin{array}{l}y\text{'}\text{'}\left(0.3\right)=2e^{0.6-0.09}\left[-1+2{(1-0.3)}^2\right]\\ \text{ =2}{e}^{0.51}(-0.02)<0\end{array}$

And

$\begin{array}{l}y\text{'}\text{'}\left(2\right)=2e^0\left[-1+2{(1-2)}^2\right]\\ \text{ =}2>0\end{array}$

Therefore, in the intervals $\left(-\infty ,1-\frac{1}{\sqrt{2}}\right)\text{ and }\left(1+\frac{1}{\sqrt{2}},\infty \right)$ function is concave up and on the interval $\left(1-\frac{1}{\sqrt{2}},1+\frac{1}{\sqrt{2}}\right)$ it is concave down.

Thus, by the above information’s graph of y is: