Explanation The second derivative test for relative extrema consists of following steps: Find all the critical points of a continuous function defined on an interval ( a , b ) . Let’s say ‘c’. x = c is a point of local maxima if f ( c ) = 0 and f ' ' ( c ) < 0 . In this case f ( c ) is then the local maximum value. x = c is a point of local minima if f ( c ) = 0 and f ' ' ( c ) > 0 . In this case f ( c ) is the local minimum value. The test fails if f ( c ) = 0 and f " (c) = 0. In this case, we go back to first derivative test. Given, f ( x ) = ( x 2 − 3 x + 1 ) e − x Differentiating function with respect to x f ' ( x ) = − e − x ( x

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Chapter 4.3, Problem 40PS

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The given critical points as relative maxima and relative minima.

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Chapter 4.3, Problem 39PS

Chapter 4.3, Problem 41PS

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Chapter 4 Solutions

Calculus

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The given critical points as relative maxima and relative minima .

Solution Summary: The author explains how the second derivative test for relative extrema consists of finding all the critical points of a continuous function defined on an interval (a,b,).

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