Explanation Given: The function is f ( x ) = ( 1000 − x ) 2 + x 2 . Concept used: If a function f is differentiable on the interval ( a , b ) then the function f is increasing on ( a , b ) , if f ′ ( x ) > 0 for x ∈ ( a , b ) and the function f is decreasing on ( a , b ) , if f ′ ( x ) < 0 for x ∈ ( a , b ) . Calculation: The given function is expressed as shown below: f ( x ) = ( 1000 − x ) 2 + x 2 …… (1) To obtain the critical point differentiate the above equation as shown below: d d x f ( x ) = d d x ( ( 1000 − x ) 2 + x 2 ) f ′ ( x ) = 2 ( 1000 − x ) ⋅ ( − 1 ) + 2 x = − ( 2000 − 2 x ) + 2 x = 4 x − 2000 To obtain the critical point equate the function f ′ ( x ) to 0. f ′ ( x ) = 0 4 x − 2000 = 0 x = 2000 4 x = 500 The critical point is 500 . Solve the inequality f ′ ( x ) < 500 as shown below: f ′ ( x ) < 0 4 x − 2000 < 0 4 x < 2000 x < 500 The function f ′ ( x ) is negative for x < 500 so, the function is decreasing in the region − ∞ < x < 500

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

4th Edition

ISBN: 9781319057664

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 4.3, Problem 65E

To determine

To find:

The interval such that the function f(x)=(1000−x)2+x2 is decreasing and find the larger one between 8002+2002 or 6002+4002.

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Chapter 4.3, Problem 64E

Chapter 4.3, Problem 66E

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Chapter 4 Solutions

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

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Prob. 63CRE Ch. 4 - Prob. 64CRE Ch. 4 - Prob. 65CRE Ch. 4 - Prob. 66CRE Ch. 4 - Prob. 67CRE Ch. 4 - Prob. 68CRE Ch. 4 - Prob. 69CRE Ch. 4 - Prob. 70CRE Ch. 4 - Prob. 71CRE Ch. 4 - Prob. 72CRE Ch. 4 - Prob. 73CRE Ch. 4 - Prob. 74CRE Ch. 4 - Prob. 75CRE Ch. 4 - Prob. 76CRE Ch. 4 - Prob. 77CRE Ch. 4 - Prob. 78CRE Ch. 4 - Prob. 79CRE Ch. 4 - Prob. 80CRE Ch. 4 - Prob. 81CRE Ch. 4 - Prob. 82CRE Ch. 4 - Prob. 83CRE Ch. 4 - Prob. 84CRE Ch. 4 - Prob. 85CRE Ch. 4 - Prob. 86CRE Ch. 4 - Prob. 87CRE Ch. 4 - Prob. 88CRE Ch. 4 - Prob. 89CRE

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The interval such that the function f ( x ) = ( 1000 − x ) 2 + x 2 is decreasing and find the larger one between 800 2 + 200 2 or 600 2 + 400 2 .

Solution Summary: The author explains that the function f(x)= (1000-x-), is decreasing and finding the larger one between 8002+2002.

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To find:

The interval such that the function f(x)=(1000−x)2+x2 is decreasing and find the larger one between 8002+2002 or 6002+4002.

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Chapter 4 Solutions

The interval such that the function f ( x ) = ( 1000 − x ) 2 + x 2 is decreasing and find the larger one between 800 2 + 200 2 or 600 2 + 400 2 .

Solution Summary: The author explains that the function f(x)= (1000-x-), is decreasing and finding the larger one between 8002+2002.

Concept explainers

Videos

To find: The interval such that the function f(x)=(1000−x)2+x2 is decreasing and find the larger one between 8002+2002 or 6002+4002.

Want to see the full answer?

Chapter 4 Solutions

To find:

The interval such that the function f(x)=(1000−x)2+x2 is decreasing and find the larger one between 8002+2002 or 6002+4002.