PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Question
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Chapter 3.1, Problem 57E

a.

To determine

Draw the graph of the function .

a.

Expert Solution
Check Mark

Explanation of Solution

Given: It is given in the question that the function is f(x)=x2e−x .

Graph:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.1, Problem 57E , 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:

b.

To determine

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

b.

Expert Solution
Check Mark

Answer to Problem 57E

The increasing and decreasing intervals for the functions are (0,2.012) and (−∞,0),(2.012,∞) .

Explanation of Solution

Given: It is given in the question that the function is f(x)=x2e−x .

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 57E , additional homework tip 2

As,it is seen in the curve cuts the x − axis at 0 and y − axis is approximately 2.012 .

So,

The interval of the increasing is (0,2.012) .

The interval of the decreasing is (−∞,0),(2.012,∞) .

Conclusion:

The increasing and decreasing intervals are (0,2.012) and (−∞,0),(2.012,∞) .

c.

To determine

Find the approximate relative maximum or minimum values.

c.

Expert Solution
Check Mark

Answer to Problem 57E

The local minima and maxima of the function are 0 and 0.542 .

Explanation of Solution

Given:

It is given in the question that the function is f(x)=x2e−x .

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)=x2e−x .

Differentiate the function,

  f(x)=x2e−xf'(x)=e−x⋅2x+x2⋅e−x⋅(−1)    =e−x(2x−x2)

Now,the interval that is found in the above part is (0,2.012) .Put the value in the first derivative.

  f'(0)=e−0(2⋅0−02)=0 (Local minima)

  f'(2.012)=e−2.012(2⋅2.012−2.0122)≈0.542 (Local Maxima)

Conclusion:

The local minima and maxima are 0 and 0.542 .

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Chapter 3.1, Problem 56E
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Chapter 3.1, Problem 58E

Chapter 3 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Prob. 32ECh. 3.6 - Prob. 33ECh. 3.6 - Prob. 34ECh. 3.6 - Prob. 35ECh. 3.6 - Prob. 36ECh. 3.6 - Prob. 37ECh. 3.6 - Prob. 38ECh. 3.6 - Prob. 39ECh. 3.6 - Prob. 40ECh. 3.6 - Prob. 41ECh. 3.6 - Prob. 42ECh. 3.6 - Prob. 43ECh. 3.6 - Prob. 44ECh. 3.6 - Prob. 45ECh. 3.6 - Prob. 46ECh. 3 - Prob. 1CRCh. 3 - Prob. 2CRCh. 3 - Prob. 3CRCh. 3 - Prob. 4CRCh. 3 - Prob. 5CRCh. 3 - Prob. 6CRCh. 3 - Prob. 7CRCh. 3 - Prob. 8CRCh. 3 - Prob. 9CRCh. 3 - Prob. 10CRCh. 3 - Prob. 11CRCh. 3 - Prob. 12CRCh. 3 - Prob. 13CRCh. 3 - Prob. 14CRCh. 3 - Prob. 15CRCh. 3 - Prob. 16CRCh. 3 - Prob. 17CRCh. 3 - Prob. 18CRCh. 3 - Prob. 19CRCh. 3 - Prob. 20CRCh. 3 - Prob. 21CRCh. 3 - Prob. 22CRCh. 3 - Prob. 23CRCh. 3 - Prob. 24CRCh. 3 - Prob. 25CRCh. 3 - Prob. 26CRCh. 3 - Prob. 27CRCh. 3 - Prob. 28CRCh. 3 - Prob. 29CRCh. 3 - Prob. 30CRCh. 3 - Prob. 31CRCh. 3 - Prob. 32CRCh. 3 - Prob. 33CRCh. 3 - Prob. 34CRCh. 3 - Prob. 35CRCh. 3 - Prob. 36CRCh. 3 - Prob. 37CRCh. 3 - Prob. 38CRCh. 3 - Prob. 39CRCh. 3 - Prob. 40CRCh. 3 - Prob. 41CRCh. 3 - Prob. 42CRCh. 3 - Prob. 43CRCh. 3 - Prob. 44CRCh. 3 - Prob. 45CRCh. 3 - Prob. 46CRCh. 3 - Prob. 47CRCh. 3 - Prob. 48CRCh. 3 - Prob. 49CRCh. 3 - Prob. 50CRCh. 3 - Prob. 51CRCh. 3 - Prob. 52CRCh. 3 - Prob. 53CRCh. 3 - Prob. 54CRCh. 3 - Prob. 55CRCh. 3 - Prob. 56CRCh. 3 - Prob. 57CRCh. 3 - Prob. 58CRCh. 3 - Prob. 59CRCh. 3 - Prob. 60CRCh. 3 - Prob. 61CRCh. 3 - Prob. 62CRCh. 3 - Prob. 63CRCh. 3 - Prob. 64CRCh. 3 - Prob. 65CRCh. 3 - Prob. 66CRCh. 3 - Prob. 67CRCh. 3 - Prob. 68CRCh. 3 - Prob. 69CRCh. 3 - Prob. 70CRCh. 3 - Prob. 71CRCh. 3 - Prob. 72CRCh. 3 - Prob. 73CRCh. 3 - Prob. 74CRCh. 3 - Prob. 75CRCh. 3 - Prob. 76CRCh. 3 - Prob. 77CRCh. 3 - Prob. 78CRCh. 3 - Prob. 79CRCh. 3 - Prob. 80CRCh. 3 - Prob. 81CRCh. 3 - Prob. 82CRCh. 3 - Prob. 83CRCh. 3 - Prob. 84CRCh. 3 - Prob. 85CRCh. 3 - Prob. 86CRCh. 3 - Prob. 87CRCh. 3 - Prob. 88CRCh. 3 - Prob. 89CRCh. 3 - Prob. 90CRCh. 3 - Prob. 91CRCh. 3 - Prob. 92CRCh. 3 - Prob. 93CRCh. 3 - Prob. 94CRCh. 3 - Prob. 95CRCh. 3 - Prob. 96CRCh. 3 - Prob. 97CRCh. 3 - Prob. 98CRCh. 3 - Prob. 99CRCh. 3 - Prob. 100CRCh. 3 - Prob. 101CRCh. 3 - Prob. 102CRCh. 3 - Prob. 103CRCh. 3 - Prob. 104CRCh. 3 - Prob. 105CRCh. 3 - Prob. 106CRCh. 3 - Prob. 107CRCh. 3 - Prob. 108CRCh. 3 - Prob. 109CRCh. 3 - Prob. 110CRCh. 3 - Prob. 111CRCh. 3 - Prob. 112CRCh. 3 - Prob. 113CRCh. 3 - Prob. 114CRCh. 3 - Prob. 115CRCh. 3 - Prob. 116CRCh. 3 - Prob. 117CRCh. 3 - Prob. 118CRCh. 3 - Prob. 119CRCh. 3 - Prob. 120CRCh. 3 - Prob. 121CRCh. 3 - Prob. 122CRCh. 3 - Prob. 123CRCh. 3 - Prob. 124CRCh. 3 - Prob. 125CRCh. 3 - Prob. 126CRCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - Prob. 16CTCh. 3 - Prob. 17CTCh. 3 - Prob. 18CTCh. 3 - Prob. 19CTCh. 3 - Prob. 20CTCh. 3 - Prob. 21CTCh. 3 - Prob. 22CTCh. 3 - Prob. 23CTCh. 3 - Prob. 24CTCh. 3 - Prob. 25CTCh. 3 - Prob. 26CTCh. 3 - Prob. 27CTCh. 3 - Prob. 28CTCh. 3 - Prob. 29CTCh. 3 - Prob. 30CTCh. 3 - Prob. 1STPCh. 3 - Prob. 2STPCh. 3 - Prob. 3STPCh. 3 - Prob. 4STPCh. 3 - Prob. 5STPCh. 3 - Prob. 6STPCh. 3 - Prob. 7STPCh. 3 - Prob. 8STPCh. 3 - Prob. 9STPCh. 3 - Prob. 10STPCh. 3 - Prob. 11STPCh. 3 - Prob. 12STPCh. 3 - Prob. 13STPCh. 3 - Prob. 14STPCh. 3 - Prob. 15STPCh. 3 - Prob. 16STPCh. 3 - Prob. 17STPCh. 3 - Prob. 18STPCh. 3 - Prob. 19STPCh. 3 - Prob. 20STPCh. 3 - Prob. 21STPCh. 3 - Prob. 22STPCh. 3 - Prob. 23STPCh. 3 - Prob. 24STPCh. 3 - Prob. 25STPCh. 3 - Prob. 26STP
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