12. Find the maximum and minimum values of  x + \sin 2x \: \:on \: \: [ 0 , 2 \pi ]

Answers (1)

Given function is
f(x) =x+ \sin 2x
f^{'}(x) =1+ 2\cos 2x\\ f^{'}(x) = 0\\ 1+2\cos 2x = 0\\ as \ x \ \epsilon \ [0,2\pi]\\ 0 < x < 2\pi\\ 0< 2x < 4\pi\\ \cos 2x = \frac{-1}{2} \ at \ 2x = 2n\pi \pm \frac{2\pi}{3} \ where \ n \ \epsilon \ Z\\ x = n\pi \pm \frac{\pi}{3}\\ x = \frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3} \ as \ x \ \epsilon \ [0,2\pi]
So, values of x are
x = \frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}  These are the critical points of the function f(x) = x+\sin 2x
Now, we need to find the value of the function f(x) = x+\sin 2x at  x = \frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}  and at the end points of given  range i.e. at x = 0 and x = 2\pi

f(x) =x+ \sin 2x\\ f(\frac{\pi}{3}) = \frac{\pi}{3}+\sin 2\left ( \frac{\pi}{3} \right ) = \frac{\pi}{3}+\sin \frac{2\pi}{3} = \frac{\pi}{3}+\frac{\sqrt3}{2}

f(x) =x+ \sin 2x\\ f(\frac{2\pi}{3}) = \frac{2\pi}{3}+\sin 2\left ( \frac{2\pi}{3} \right ) = \frac{2\pi}{3}+\sin \frac{4\pi}{3} = \frac{2\pi}{3}-\frac{\sqrt3}{2}

f(x) =x+ \sin 2x\\ f(\frac{4\pi}{3}) = \frac{4\pi}{3}+\sin 2\left ( \frac{4\pi}{3} \right ) = \frac{4\pi}{3}+\sin \frac{8\pi}{3} = \frac{4\pi}{3}+\frac{\sqrt3}{2}

f(x) =x+ \sin 2x\\ f(\frac{5\pi}{3}) = \frac{5\pi}{3}+\sin 2\left ( \frac{5\pi}{3} \right ) = \frac{5\pi}{3}+\sin \frac{10\pi}{3} = \frac{5\pi}{3}-\frac{\sqrt3}{2}

f(x) = x+\sin 2x\\ f(2\pi) = 2\pi+\sin 2(2\pi)= 2\pi+\sin 4\pi = 2\pi

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

Hence, at   x = 2\pi function f(x) = x+\sin 2x attains its maximum value and value is 2\pi in the given range of x \ \epsilon \ [0,2\pi]
and at x= 0 function f(x) = x+\sin 2x attains its minimum value and value is 0

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