Explanation The first 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’. If f ' ( x ) changes sign from positive to negative as x increases through c, then c is a point of local maxima, and f ( c ) is local maximum value. If f ' ( x ) changes sign from negative to positive as x increases through c, then c is a point of local minima, and f f ( c ) is local minimum value. If f ' ( x ) does not change sign as x increases through c, then c is neither a point of local minima nor a point of local maxima. Such a point is called a point of inflection. Given, f ( x ) = x 3 − 48 x 3 Now differentiating with respect to ‘x’. f ' ( x ) = x 2 − 16 ( x 3 − 48 x ) 2 3 Let f ' ( x ) = 0

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Chapter 4.3, Problem 37PS

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Relative maxima and relative minima using first derivative test.

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Chapter 4.3, Problem 36PS

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1. Use the Second Derivative Test to find the Maximum and Minimum value(s) of the function 3 f(x) = x - 4x+ Inx. Give exact solutions. 2

9. In our last class discussion we found that a function y(x) = x* indicated by taking y"(x) = 0 that there should be an inflection at x = 0, but the first derivative test identified x = 0 as a relative minimum. What term did we use to classify functions with this behavior?

3. Find the stationary points of the function f(x) = 3.x* – 2xº. Apply the Second Derivative Test to identify them as local minimum, local maximum or undetermined. If the Second Derivative Test did not help, use the First Derivative Test instead.

Chapter 4 Solutions

Calculus

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Relative maxima and relative minima using first derivative test.

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

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