8. At what points in the interval [ 0 , 2 \pi ] does the function \sin 2x  attain its maximum
value?

Answers (1)

Given function is
f(x) = \sin 2x
f^{'}(x) = 2\cos 2x\\ f^{'}(x) = 0\\ 2\cos 2x = 0\\ as \ x \ \epsilon [0,2\pi]\\ 0 < x < 2\pi\\ 0< 2x < 4\pi\\ \cos 2x = 0 \ at \ 2x = \frac{\pi}{2},2x = \frac{3\pi}{2},2x=\frac{5\pi}{2}and 2x= \frac{7\pi}{2}\\
So, values of x are
x = \frac{\pi}{4},x = \frac{3\pi}{4},x=\frac{5\pi}{4} \ and \ x= \frac{7\pi}{4}\\  These are the critical points of the function f(x) = \sin 2x
Now, we need to find the value of the function f(x) = \sin 2x at x = \frac{\pi}{4},x = \frac{3\pi}{4},x=\frac{5\pi}{4} \ and \ x= \frac{7\pi}{4}\\  and at the end points of given  range i.e. at x = 0 and x = \pi

f(x) = \sin 2x\\ f(\frac{\pi}{4}) = \sin 2\left ( \frac{\pi}{4} \right ) = \sin \frac{\pi}{2} = 1

f(x) = \sin 2x\\ f(\frac{3\pi}{4}) = \sin 2\left ( \frac{3\pi}{4} \right ) = \sin \frac{3\pi}{2} = -1

f(x) = \sin 2x\\ f(\frac{5\pi}{4}) = \sin 2\left ( \frac{5\pi}{4} \right ) = \sin \frac{5\pi}{2} = 1

f(x) = \sin 2x\\ f(\frac{7\pi}{4}) = \sin 2\left ( \frac{7\pi}{4} \right ) = \sin \frac{7\pi}{2} = -1

f(x) = \sin 2x\\ f(\pi) = \sin 2(\pi)= \sin 2\pi = 0

f(x) = \sin 2x\\ f(0) = \sin 2(0)= \sin 0 = 0

Hence, at  x =\frac{\pi}{4} \ and \ x = \frac{5\pi}{4} function f(x) = \sin 2x attains its maximum value i.e. in 1 in the given range of x \ \epsilon \ [0,2\pi]
 

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