Browse Questions

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

Toolbox:
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} • \large\frac{d}{dx}$$(\sin x)=\cos x$
Step 1:
Let $f(x)=x+\sin 2x$
$f'(x)=1+2\cos 2x$
For maxima and minima
$f'(x)=0$
$1+2\cos 2x=0$
$2\cos 2x=-1$
$\cos 2x=\large\frac{-1}{2}$
$2x=\cos^{-1}\big(\large\frac{-1}{2}\big)$
$\quad=\large\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3},\frac{10\pi}{3}$
$x=\large\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
Step 2:
Now we find $f(x)$ at $x=0,\large\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3},$$2\pi We have f'(x)=1+2\cos 2x f(x)=x+\sin 2x f(0)=0+\sin 0=0 f(\large\frac{\pi}{3})=\large\frac{\pi}{3}+$$\sin\large\frac{2\pi}{3}$
$\qquad=\large\frac{\pi}{3}+\frac{\sqrt 3}{2}$
$f(\large\frac{\pi}{3})=\large\frac{2\pi}{3}+$$\sin\large\frac{4\pi}{3} \qquad=\large\frac{2\pi}{3}+\frac{\sqrt 3}{2} f(\large\frac{4\pi}{3})=\large\frac{4\pi}{3}+$$\sin\large\frac{8\pi}{3}$
$\qquad=\large\frac{4\pi}{3}+\frac{\sqrt 3}{2}$
$f(\large\frac{5\pi}{3})=\large\frac{5\pi}{3}+$$\sin\large\frac{10\pi}{3}$
$\qquad=\large\frac{5\pi}{3}+\frac{\sqrt 3}{2}$
$f(2\pi)=2\pi+\sin 2\pi$
$\qquad\;=2\pi$
Hence maximum $f(x)=2\pi$
Minimum $f(x)=0$

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