Explanation Theorem used: “Suppose f is continuous on an interval I and differentiable at all interior points of I . If f ′ ( x ) > 0 at all interior points of I , then f is increasing on I . If f ′ ( x ) < 0 at all interior points of I , then f is decreasing in I ”. Definition used: “An interior point c of the domain of f at which f ′ ( c ) = 0 or f ′ ( c ) fails to exist is called critical point of f ”. Calculation: Obtain the first derivative of the function f ( x ) . f ′ ( x ) = d d x ( x 2 − 16 ) = d d x ( x 2 ) − d d x ( 16 ) = 2 x − 0 = 2 x That is, f ′ ( x ) = 2 x . Equate f ′ ( x ) = 0 and obtain the critical point. 2 x = 0 x = 0 Therefore, the critical point is x = 0 . The critical point divide the interval ( − ∞ , ∞ ) into ( − ∞ , 0 ) and ( 0 , ∞ )

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

3rd Edition

ISBN: 9780134996103

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 4.3, Problem 20E

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To find: The intervals on which f(x)=x2−16 is increasing and the intervals in which f(x) is decreasing.

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Chapter 4.3, Problem 19E

Chapter 4.3, Problem 21E

Chapter 4 Solutions

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

Ch. 4.1 - Sketch the graph of a function that is continuous...Ch. 4.1 - Consider the function f(x) = x3. Where is the...Ch. 4.1 - Sketch the graph of a function that is continuous...Ch. 4.1 - Sketch the graph of a function that has an...Ch. 4.1 - What is a critical point of a function?Ch. 4.1 - Sketch the graph of a function f that has a local...Ch. 4.1 - Sketch the graph of a function f that has a local...Ch. 4.1 - Absolute maximum/minimum values Use the following...Ch. 4.1 - Prob. 12E Ch. 4.1 - Absolute maximum/minimum values Use the following...

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The intervals on which f ( x ) = x 2 − 16 is increasing and the intervals in which f ( x ) is decreasing.

Solution Summary: The author explains how the function f(x)=x 2-16 is decreasing and increasing on the interval.

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To find: The intervals on which f(x)=x2−16 is increasing and the intervals in which f(x) is decreasing.

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Chapter 4 Solutions