Question

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

Answer 1

Let $f\left(x\right)=sin2x\Rightarrow f^{\prime }\left(x\right)=2cos2x$
For maxima or minima put $f^{\prime }\left(x\right)=0$
$\Rightarrow 2cos2x=0\Rightarrow cos2x=0\Rightarrow 2x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{\pi }{2},\frac{7\pi }{2}$
Thus, 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 end points of the interval $[0,2\pi ]$.
$\begin{array}{cc}Atx=0& f\left(0\right)=sin(2\times 0)=0\\ Atx=2\pi ,& f\left(2\pi \right)=sin(2\times 2\pi )=0\\ Atx=\frac{\pi }{4},& f\left(\frac{\pi }{4}\right)=sin(2\times \frac{\pi }{4})=sin\frac{\pi }{2}=1\\ Atx=\frac{3\pi }{4}& f\left(\frac{3\pi }{4}\right)=sin(2\times \frac{3\pi }{4})=sin\frac{3\pi }{2}=sin(\pi +\frac{\pi }{2})=-sin\frac{\pi }{2}=-1\\ Atx=\frac{5\pi }{4}& f\left(\frac{5\pi }{4}\right)=sin(2\times \frac{5\pi }{4})=sin\frac{5\pi }{2}=sin(2\pi +\frac{\pi }{2})=sin\frac{\pi }{2}=1\\ Atx=\frac{7\pi }{4}& f\left(\frac{7\pi }{4}\right)=sin(2\times \frac{7\pi }{4})=sin\frac{7\pi }{2}=sin(2\pi +\frac{3\pi }{2})=sin\frac{3\pi }{2}=-1\end{array}$
Hence, we can conclude that absolute maximum value of f on $[0,2\pi ]$ is 1 occuring at $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$