PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

8th Edition

ISBN: 9781337904285

Author: Larson

Publisher: CENGAGE L

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Question

Chapter 3.1, Problem 58E

To determine

Draw the graph of the function .

Expert Solution

Explanation of Solution

Given: It is given in the question that the function is f(x)=2x2ex+1 .

Graph:

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.1, Problem 58E , additional homework tip 1

Interpretation: The graph for the function f(x)=x2e−x is given above.

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

Conclusion:

To determine

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

Expert Solution

Answer to Problem 58E

The increasing and decreasing intervals for the fumction are (−∞,−2.125),(0,∞) and (−2.125,0) .

Explanation of Solution

Given:

It is given in the question that the function is f(x)=2x2ex+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:

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.1, Problem 58E , additional homework tip 2

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 (−∞,−2.125),(0,∞) .

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

Conclusion:

The increasing and decreasing intervals are (−∞,−2.125),(0,∞) and (−2.125,0) .

To determine

Find the approximate relative maximum or minimum values.

Expert Solution

Answer to Problem 58E

The Maximum +∞ 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)=2x2ex+1 .

Concept Used:

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

Calculation: The function is f(x)=2x2ex+1 .

Differentiate the function,

f(x)=2x2ex+1f'(x)=2(ex+1⋅2x+x2⋅ex+1⋅(1)) =2ex+1(2x+x2)

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

f'(0)=2e0+1(2⋅0+02)=0 (Local minima)

f'(−2.125)=2e−2.125+1(2⋅−2.125+2.1252)≈2.95 (Local Maxima),Maximum: +∞

Conclusion:

The Maximum +∞ is and local minima and maxima are 0 and 2.95 .

Chapter 3.1, Problem 57E

Chapter 3.1, Problem 59E

Chapter 3 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Prob. 111CR Ch. 3 - Prob. 112CR Ch. 3 - Prob. 113CR Ch. 3 - Prob. 114CR Ch. 3 - Prob. 115CR Ch. 3 - Prob. 116CR Ch. 3 - Prob. 117CR Ch. 3 - Prob. 118CR Ch. 3 - Prob. 119CR Ch. 3 - Prob. 120CR Ch. 3 - Prob. 121CR Ch. 3 - Prob. 122CR Ch. 3 - Prob. 123CR Ch. 3 - Prob. 124CR Ch. 3 - Prob. 125CR Ch. 3 - Prob. 126CR Ch. 3 - Prob. 1CT Ch. 3 - Prob. 2CT Ch. 3 - Prob. 3CT Ch. 3 - Prob. 4CT Ch. 3 - Prob. 5CT Ch. 3 - Prob. 6CT Ch. 3 - Prob. 7CT Ch. 3 - Prob. 8CT Ch. 3 - Prob. 9CT Ch. 3 - Prob. 10CT Ch. 3 - Prob. 11CT Ch. 3 - Prob. 12CT Ch. 3 - Prob. 13CT Ch. 3 - Prob. 14CT Ch. 3 - Prob. 15CT Ch. 3 - Prob. 16CT Ch. 3 - Prob. 17CT Ch. 3 - Prob. 18CT Ch. 3 - Prob. 19CT Ch. 3 - Prob. 20CT Ch. 3 - Prob. 21CT Ch. 3 - Prob. 22CT Ch. 3 - Prob. 23CT Ch. 3 - Prob. 24CT Ch. 3 - Prob. 25CT Ch. 3 - Prob. 26CT Ch. 3 - Prob. 27CT Ch. 3 - Prob. 28CT Ch. 3 - Prob. 29CT Ch. 3 - Prob. 30CT Ch. 3 - Prob. 1STP Ch. 3 - Prob. 2STP Ch. 3 - Prob. 3STP Ch. 3 - Prob. 4STP Ch. 3 - Prob. 5STP Ch. 3 - Prob. 6STP Ch. 3 - Prob. 7STP Ch. 3 - Prob. 8STP Ch. 3 - Prob. 9STP Ch. 3 - Prob. 10STP Ch. 3 - Prob. 11STP Ch. 3 - Prob. 12STP Ch. 3 - Prob. 13STP Ch. 3 - Prob. 14STP Ch. 3 - Prob. 15STP Ch. 3 - Prob. 16STP Ch. 3 - Prob. 17STP Ch. 3 - Prob. 18STP Ch. 3 - Prob. 19STP Ch. 3 - Prob. 20STP Ch. 3 - Prob. 21STP Ch. 3 - Prob. 22STP Ch. 3 - Prob. 23STP Ch. 3 - Prob. 24STP Ch. 3 - Prob. 25STP Ch. 3 - Prob. 26STP

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