Explanation Given: The function is shown below. y = ( 1 − x ) e a x ...... (1) Calculation: The given function is an exponential function. To obtain the critical point of the function, differentiate the equation (1) as shown below: d y d x = d d x ( 1 − x ) e a x y ′ = ( 1 − x ) d d x e a x + e a x d d x ( 1 − x ) y ′ = ( 1 − x ) a e a x + e a x ⋅ ( − 1 ) y ′ = a e a x ( 1 − x ) − e a x To obtain the critical point equate the function y ′ to 0 . a e a x ( 1 − x ) − e a x = 0 e a x ( a ( 1 − x ) ) = 0 The value of e a x cannot be equal to zero. Therefore, by zero product property the value of x is calculated as shown below: 1 − x = 0 x = 1 The critical point is x = 1

4th Edition

ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 4.2, Problem 84E

To determine

To find:

The maximum and minimum value of the function y=(1−x)eax on the interval [0,1] and 0<a.

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Chapter 4.2, Problem 83E

Chapter 4.2, Problem 85E

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(#1) Consider the solid bounded below by z = x² and above by z = 4-y². If we were to project this solid down onto the xy-plane, you should be able to use algebra to determine the 2D region R in the xy-plane for the purposes of integration. Which ONE of these limite of integration would correctly describe R? (a) y: x24x: -22 - (b) y: 22 x: 04-y² (c) y: -√√4-x2. →√√4x²x: −2 → 2 (d) z: 24-y² y: -2 → 2 (e) None of the above

X MindTap - Cenxxxx Answered: tat "X A 26308049 X 10 EKU-- SP 25: X E DNA Sequenc X b/ui/evo/index.html?elSBN=9780357038406&id=339416021&snapshotid=877369& GE MINDTAP , Limits, and the Derivative 40. Answer 5 4-5 t-10 5 f(x) = 2x - 4 if x ≤0 if x 0 10 ++ -4-3-2-1 f(x) = MacBook Pro Search or type URL 5 1234 x² +1 if x = 0 if x = 0 +

Chapter 4 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

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The maximum and minimum value of the function y = ( 1 − x ) e a x on the interval [ 0 , 1 ] and 0 < a .

Solution Summary: The author explains that the given function is an exponential function. To obtain the critical point of the function, differentiate the equation as shown below.

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To find:

The maximum and minimum value of the function y=(1−x)eax on the interval [0,1] and 0<a.

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Chapter 4 Solutions

The maximum and minimum value of the function y = ( 1 − x ) e a x on the interval [ 0 , 1 ] and 0 &lt; a .

Solution Summary: The author explains that the given function is an exponential function. To obtain the critical point of the function, differentiate the equation as shown below.

Concept explainers

Videos

To find: The maximum and minimum value of the function y=(1−x)eax on the interval [0,1] and 0<a.

Want to see the full answer?

Chapter 4 Solutions

The maximum and minimum value of the function y = ( 1 − x ) e a x on the interval [ 0 , 1 ] and 0 < a .

To find:

The maximum and minimum value of the function y=(1−x)eax on the interval [0,1] and 0<a.