Browse Questions

# Find all the local maximum values and local minimum values of the function $f(x)=\sin 2x-x, -\large\frac{\pi}{2} < x < \large\frac{\pi}{2}$

Toolbox:
• $f$ is said to have a maximum value in $I$ , if there exist a point c in I such that $f(c) \geq f (x)$ for all $x \in I$.The number $f( c)$ is called the maximum value of f in I and the point c is called a point of maximum value of f in I
• $f$ is said to have a minimum value in $I$ , if there exist a point $c$ in I such that $f(c) \leq f (x)$ for all $x \in I$.The number $f(c)$ is called the minimum value of f in I and the point $c$ in this case is called a point of minimum value of $f$ in I
• $f$ is said to have a extreme value in $I$ , if there exist a point $c$ in I such that f(c) is either a maximum value or minimum value of $f$ in $I$. The number $f (c)$ in this case is called the extreme value of $f$ in $I$ and the point $c$ is called the extreme point.
Step 1:
Let $f(x)=\sin 2x-x,\large\frac{-\pi}{2}$$\leq x\leq \large\frac{\pi}{2} Differentiating w.r.t x f'(x)=2\cos 2x-1=0 \cos 2x=\large\frac{1}{2} \cos 2x=\cos\large\frac{\pi}{3}$$,\cos(-\large\frac{\pi}{3})$
$\Rightarrow 2x=\large\frac{\pi}{3},-\frac{\pi}{3}$
$x=\large\frac{\pi}{6},\frac{-\pi}{6}$
Step 2:
Consider $x=-\large\frac{\pi}{6}$
Case (i) when $x < -\large\frac{\pi}{6}\Rightarrow$$2x <-\large\frac{\pi}{3} -2x < \large\frac{\pi}{3} \cos (-2x) < \cos\large\frac{\pi}{3} (\cos x is decreasing when 0 < x < \pi) \cos 2x < \large\frac{1}{2} \Rightarrow 2\cos 2x < 1 2\cos 2x-1 < 0 f'(x) < 0 Step 3: Case (ii) when x > \large\frac{\pi}{6} -2x > \large\frac{\pi}{3} \cos (-2x) > \cos\large\frac{\pi}{3} (\cos x is decreasing when 0 < x < \pi) \cos 2x > \large\frac{1}{2} \Rightarrow 2\cos 2x > 1 2\cos 2x-1 > 0 f'(x) > 0 Since f'(x) changes sign from -ve to +ve. When it passes through x=-\large\frac{\pi}{6} x=-\large\frac{\pi}{6} is a point of local minima. Local minimum value f(-\large\frac{\pi}{6}) \Rightarrow \sin 2(-\large\frac{\pi}{6})-(\large\frac{-\pi}{6}) \Rightarrow -\sin\large\frac{\pi}{3}+\frac{\pi}{6} \Rightarrow \large\frac{\sqrt 3}{2}+\frac{\pi}{6} \Rightarrow \large\frac{\pi-3\sqrt 3}{6} Step 4: Consider x=\large\frac{\pi}{6} Case (i):When x < \large\frac{\pi}{6} \Rightarrow 2x < \large\frac{\pi}{3} \cos 2x > \large\frac{1}{2} (\cos x decreasing when 0 < x < \pi) 2\cos 2x <1\Rightarrow 2\cos 2x-1 <0 f'(x) >0 Step 5: Case (ii) When x > \large\frac{\pi}{6} \Rightarrow 2x > \large\frac{\pi}{3} \cos 2x < \large\frac{1}{2} (\cos x decreasing when 0 < x < \pi) 2\cos 2x <1\Rightarrow 2\cos 2x-1 <0 f'(x) >0 Since f'(x) changes sign from +ve to -ve when it passes through x=\large\frac{\pi}{6} \therefore x=\large\frac{\pi}{6} is a point of local maxima. Local maximum value f(\large\frac{\pi}{6})$$=\sin 2(\large\frac{\pi}{6})-\frac{\pi}{6}$
$\Rightarrow \sin\large\frac{\pi}{3}-\frac{\pi}{6}$
$\Rightarrow \large\frac{\sqrt 3}{2}-\frac{\pi}{6}$