Question

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

Answer 1

Let f(x) = sin 2x

f'(x)=2cos2x

Now,

f'(x)=0=cos2x=0

2x=$\frac{\pi}{2}$,$\frac{3\pi}{2}$,$\frac{5\pi}{2}$,$\frac{7\pi}{4}$

x=$\frac{\pi}{4}$,$\frac{3\pi}{4}$,$\frac{5\pi}{4}$,$\frac{7\pi}{4}$

Then, we evaluate the values of f at critical points x=$\frac{\pi}{4}$,$\frac{3\pi}{4}$,$\frac{5\pi}{4}$,$\frac{7\pi}{4}$ and at the endpoints of the interval [0, 2$\pi$].

f($\frac{\pi}{4}$)=sin $\frac{\pi}{2}$=1.f($\frac{3\pi}{2}$)=$\frac{3\pi}{2}$=-1

f($\frac{5\pi}{4}$)=sin $\frac{5\pi}{2}$=1.f($\frac{7\pi}{4}$)=sin $\frac{7\pi}{2}$=-1

f(0)=sin 0=0,f(2$\pi$)=sin 2$\pi$=0

Hence, we can conclude that the absolute maximum value of f on [0, 2$\pi$] is occurring

at x=$\frac{\pi}{4}$ and x=$\frac{5\pi}{4}$.