Explanation Given information: The function is f ( x ) = 1 3 − 2 e x . Formula Used: Let the function f and f ′ are differentiable on an interval L. If f ″ is positive, then f is concave up. If f ″ is negative, then f is concave down. Let the function f and f ′ are differentiable. Also, Let c be a critical point of a function f with f ′ ( c ) = 0 . Then, f ″ is continuous on an interval. If f ″ ( c ) is positive, then f has a local minimum at x = c . If f ″ ( c ) is negative, then f has a local maximum at x = c . If f ″ ( c ) is 0, then test fails. Calculation: The function is f ( x ) = 1 3 − 2 e x . Find the derivative of the function. f ′ ( x ) = d d x ( 1 3 − 2 e x ) = ( 3 − 2 e x ) d d x ( 1 ) − ( 1 ) d d x ( 3 − 2 e x

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ISBN: 9781464140051

Author: TAALMAN, Laura, Kohn, Peter

Publisher: W. H. Freeman

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Chapter 3.3, Problem 33S

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Find the local extrema of function f by the second-derivative test. If it fails, then use the first-derivative test. Also, draw the graph.

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Chapter 3.3, Problem 32S

Chapter 3.3, Problem 34S

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Chapter 3 Solutions

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Find the local extrema of function f by the second-derivative test. If it fails, then use the first-derivative test. Also, draw the graph.

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Find the local extrema of function f by the second-derivative test. If it fails, then use the first-derivative test. Also, draw the graph.

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Chapter 3 Solutions