Explanation Given Information: The provided trigonometric function is y = 3 2 x + cos x over the interval [ 0 , 2 π ] . Formula used: Cosine derivative Rule: d d x [ cos u ] = − sin u d u d x General power derivative rule: d d x x n = n x n − 1 Condition for critical points: d y d x = f ′ ( x ) = 0 If function f ( x ) is increasing, then f ′ ( x ) > 0 and if function is decreasing, then f ′ ( x ) < 0 . From first derivative Test, if f ′ ( x ) changes positive to negative, then function f ( x ) has relative maxima and, if f ′ ( x ) changes negative to positive then function f ( x ) has relative minima. In order words, if function f ( x ) changes from increasing function to decreasing at any point critical, then function f ( x ) will have relative maxima at that point, and if function f ( x ) changes from decreasing function to increasing at any critical point, then function f ( x ) will have relative minima at that point. Calculation: Consider the provided trigonometric function: y = 3 2 x + cos x Differentiate the above function with respect to x using Cosine and Power derivative rules. d y d x = d d x [ 3 2 x + cos x ] y ′ ( x ) = 3 2 d d x [ x ] + d d x [ cos x ] = 3 2 [ 1 ] + [ − sin x ] = 3 2 − sin x Apply d y d x = 0 in order to find critical points. 3 2 − sin x = 0 − sin x = − 3 2 sin x = 3 2 Since, the value of the sin θ is zero at θ = π 3 , and θ = 2 π 3 . So, sin x = 3 2 sin x = sin π 3 x = π 3 = 1.04719755 , And, sin x = 3 2 sin x = sin 2 π 3 x = 2 π 3 = 2.09439510 Thus, critical points are x = π 3 and x = 2 π 3 . Now, rewrite the first derivative of the function. y ′ ( x ) = 3 2 − sin x At x = π 4 . y ′ ( π 4 ) = 3 2 − sin π 4 = 0.1589 = + v e > 0 At x = π 2 . y ′ ( π 2 ) = 3 2 − sin π 2 = − 0.134 = − v e < 0 At x = π . y ′ ( π ) = 3 2 − sin π = 0.866 = + v e > 0 Thus, function y = 3 2 x + cos x over the interval [ 0 , 2 π ] can be defined in tabular form as: Interval 0 < x < π 3 π 3 < x < 2 π 3 2 π 3 < x < 2 π Sign of y ′ ( x ) y ′ ( x ) > 0 y ′ ( x ) < 0 y ′ ( x ) > 0 y Increasing Decreasing Increasing From above table, y ′ ( x ) is changing from positive to negative at x = π 3

Calculus: An Applied Approach (MindTap Course List)

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ISBN: 9781305860919

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 8.4, Problem 56E

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To calculate: The relative extrema of the trigonometric function y=32x+cosx over the interval [0,2π].

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Chapter 8.4, Problem 55E

Chapter 8.4, Problem 57E

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The relative extrema of the trigonometric function y = 3 2 x + cos x over the interval [ 0 , 2 π ] .

Solution Summary: The author calculates the relative extrema of the trigonometric function y=sqrt32x+mathrmcos

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To calculate: The relative extrema of the trigonometric function y=32x+cosx over the interval [0,2π].

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