Explanation Given Information: The provided trigonometric function is y = 1 2 sec x over the interval [ 0 , 2 π ] . Formula used: Secant derivative Rule: d d x [ sec u ] = sec u tan 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 = 1 2 sec x Differentiate the above function with respect to x using Secant and Power derivative rules. d y d x = d d x [ 1 2 sec x ] y ′ ( x ) = 1 2 d d x [ sec x ] = 1 2 [ sec x tan x ] = 1 2 sec x tan x Apply d y d x = 0 in order to find critical points. 1 2 sec x tan x = 0 sec x tan x = 0 Further solve. tan x = 0 And, sec x ≠ 0 Since, the value of the tan θ is zero at θ = π between [ 0 , 2 π ] . So, tan x = 0 tan x = tan π x = π = 3.141592654 Thus, critical point is x = π between [ 0 , 2 π ] . Now, rewrite the first derivative of the function. y ′ ( x ) = 1 2 sec x tan x At x = π 4 . y ′ ( π 4 ) = 1 2 sec π 4 tan π 4 = 1 2 ( 2 ) ( 1 ) = 1 2 = + v e > 0 At x = 5 π 4

Calculus: An Applied Approach (MindTap Course List)

10th Edition

ISBN: 9781305860919

Author: Ron Larson

Publisher: Cengage Learning

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Fundamentals of Trigonometric Identities

Trigonometric Identities are the equalities that involve trigonometry functions and hold for all the values of variables given in the equation.

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Chapter 8.4, Problem 58E

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To calculate: The relative extrema of the trigonometric function y=12secx over the interval [0,2π].

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Chapter 8.4, Problem 57E

Chapter 8.4, Problem 59E

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Calculus: An Applied Approach (MindTap Course List)

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The relative extrema of the trigonometric function y = 1 2 sec x over the interval [ 0 , 2 π ] .

Solution Summary: The author calculates the relative extrema of the trigonometric function y=12mathrmsecx over the interval

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To calculate: The relative extrema of the trigonometric function y=12secx over the interval [0,2π].

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Chapter 8 Solutions