Find the points of local maxima or local minima, functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be f (x) = sin 2x, 0 < x < π

Asked by Sakshi | 1 year ago |  94

##### Solution :-

Given f (x) = sin 2x

Differentiate w.r.t x, we get

f'(x) = 2cos 2x, 0 < x < π

For the point of local maxima and minima, f’(x) = 0

2cos 2x = 0

cos 2x = 0

2x = $$\dfrac{\pi}{2},\dfrac{3\pi}{2}$$

x =$$\dfrac{\pi}{4},\dfrac{3\pi}{4}$$

Now, at x = $$\dfrac{\pi}{4}$$, f’(x) changes from positive to negative

Hence, local max value $$f^{(\dfrac{\pi}{4})}=1$$

Hence, local min value $$f^{(\dfrac{3\pi}{4})}=-1$$

Answered by Sakshi | 1 year ago

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