Chapter 3.1, Problem 62E

a.

To determine

Draw the graph of the function .

Explanation of Solution

Given: It is given in the question that the function is $f\left(x\right)=2x^2{e}^{x+1}$ .

Graph:

  

Interpretation: The graph for the function $f(x)=x^2{e}^{-x}$ is given above.

In the graphic utility,simply put the function on that and the graph will get it.

Conclusion:

b.

To determine

Find the interval on which the function is increasing and decreasing.

Answer to Problem 62E

The increasing and decreasing intervals for the fumction are $(-\infty ,-2.125),(0,\infty )$ and $(-2.125,0)$ .

Explanation of Solution

Given:

It is given in the question that the function is $f(x)=2x^2{e}^{x+1}$ .

Concept Used:

In this,use the concept of understanding the graph and know to find the point of intersection.

Calculation:

As ,it is find that the graph is given below:

  

As,it is seen in the curve the two coordinates are found $(0,0)$ and $(-2.125,3.145)$ .

So,

The interval of the increasing is $(-\infty ,-2.125),(0,\infty )$ .

The interval of the decreasing is $(-2.125,0)$ .

Conclusion:

The increasing and decreasing intervals are $(-\infty ,-2.125),(0,\infty )$ and $(-2.125,0)$ .

c.

To determine

Find the approximate relative maximum or minimum values.

Answer to Problem 62E

The Maximum $+\infty$ is and local minima and maxima of the function are $0$ and $2.95$ .

Explanation of Solution

Given:

It is given in the question that the function is $f(x)=2x^2{e}^{x+1}$ .

Concept Used:

In this,use the concept of finding the maxima and minima use the $f\text{'}(x)=0$ put the values on that and get the value.

Calculation: The function is $f(x)=2x^2{e}^{x+1}$ .

Differentiate the function,

  $\begin{array}{l}f(x)=2x^2{e}^{x+1}\\ f\text{'}(x)=2(e^{x+1}\cdot 2x+x^2\cdot e^{x+1}\cdot (1))\\ =2e^{x+1}(2x+x^2)\end{array}$

Now,the interval that is found in the above part is $(-2.125,0)$ .Put the value in the first derivative.

  $f\text{'}(0)=2e^{0+1}(2\cdot 0+0^2)=0$ (Local minima)

  $f\text{'}(-2.125)=2e^{-2.125+1}(2\cdot -2.125+{2.125}^2)\approx 2.95$ (Local Maxima),Maximum: $+\infty$

Conclusion:

The Maximum $+\infty$ is and local minima and maxima are $0$ and $2.95$ .