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=\(\frac{5\pi}{4}\).