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How many extreme values does sin4x + 2x, where \(0 < x < \frac{\pi}{2} \) have ?

A.
1
B.
2
C.
4
D.
8

Solution:

Concept:First Order Derivative:Lets say we have a function f which is continuous at the critical point, defined in an open interval I andf(c) = 0 (slope is 0 at c). Then we check the value of f'(x) at the point left to the curve and right to thecurve and check the nature of f'(x), then we can say, that the given point will be:Local maxima: If f(x) changes sign from positive to negative as x increases via point c, then f(c) givesthe maximum value of the function in that range.Local minima: If f(x) changes sign from negative to positive as x increases via point c, then f(c) givesthe minimum value of the function in that range.Maxima and minima are points where a function reaches a highest or lowest value, respectively. Thesemaxima and minima of the function together are called extremes.Calculation;Let f(x) = sin4x + 2x \(0 < x < \frac{\pi}{2} \)? f?(x) = 4 cos4x + 2? f?(x) = 0 ? 4cos4x + 2 = 0? cos4x = \(\displaystyle\frac{-1}{2}\)?? 4x = \(\displaystyle \frac{2\pi}{3}\), ?\(\displaystyle \frac{4\pi}{3}\)?? x = ?\(\displaystyle \frac{\pi}{6}\), \(\displaystyle \frac{\pi}{3}\)? sin4x + 2x have 2 extreme values where \(0 < x < \frac{\pi}{2} \)

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