Question

Find the maximum and minimum values of $ x + \sin 2x $ on $ \left[ {0,2\pi } \right] $

Answer 1

Hint: Consider the given expression as a function and then find the differentiation of that function to get the value of x. And get the values of x within the given range. After getting the values of x, substitute them in the given expression and get its maximum and minimum value.

Complete step-by-step answer:
We are given an expression $ x + \sin 2x $ and we have to find its maximum value and minimum value within the range $ \left[ {0,2\pi } \right] $
 $ x + \sin 2x $ is in terms of the variable x.
Let $ x + \sin 2x $ be a function $ f\left( x \right) $
 $ f\left( x \right) = x + \sin 2x $
Now we are finding the differentiation of the function $ f\left( x \right) $ with respect to x to get the value of x.
 $
\Rightarrow \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {x + \sin 2x} \right) \\
  \Rightarrow {f^1}\left( x \right) = \dfrac{d}{{dx}}x + \dfrac{d}{{dx}}\sin 2x \\
  \Rightarrow {f^1}\left( x \right) = 1 + 2\cos 2x \\
  \left( {\because \dfrac{{dx}}{{dx}} = 1,\dfrac{d}{{dx}}\left( {\sin nx} \right) = n\cos nx} \right) \\
  $
Where $ {f^1}\left( x \right) $ is the derivative function of $ f\left( x \right) $
When the value of $ {f^1}\left( x \right) $ is 0 then the value of x will be
 $
  {f^1}\left( x \right) = 1 + 2\cos 2x = 0 \\
  \Rightarrow 1 + 2\cos 2x = 0 \\
  \Rightarrow 2\cos 2x = - 1 \\
  \Rightarrow \cos 2x = \dfrac{{ - 1}}{2} \\
  $
The value of cosine function is $ \dfrac{1}{2} $ only when the angle is 60 degrees or $ \dfrac{\pi }{3} $
 $
  \Rightarrow \cos 2x = - \cos \dfrac{\pi }{3} \\
  \Rightarrow - \cos \dfrac{\pi }{3} = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = \cos \dfrac{{2\pi }}{3} \\
  \Rightarrow \cos 2x = \cos \dfrac{{2\pi }}{3} \\
  \Rightarrow 2x = 2n\pi \pm \dfrac{{2\pi }}{3},n \in Z \\
  \Rightarrow x = n\pi \pm \dfrac{{2\pi }}{3},n \in Z \\
  $
We have got the expression to calculate the value of x, and the values of x must be in the range $ \left[ {0,2\pi } \right] $
So the values of x will be $ 0,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3},2\pi $
Now, substitute the values of x in the function $ f\left( x \right) = x + \sin 2x $ to find the function values.
 $
  x = 0,f\left( 0 \right) = 0 + \sin 2\left( 0 \right) = 0 \\
  x = \dfrac{\pi }{3},f\left( {\dfrac{\pi }{3}} \right) = \dfrac{\pi }{3} + \sin 2\dfrac{\pi }{3} = \dfrac{\pi }{3} + \dfrac{{\sqrt 3 }}{2} \\
  x = \dfrac{{2\pi }}{3},f\left( {\dfrac{{2\pi }}{3}} \right) = \dfrac{{2\pi }}{3} + \sin 2\left( {\dfrac{{2\pi }}{3}} \right) = \dfrac{{2\pi }}{3} - \dfrac{{\sqrt 3 }}{2} \\
  x = \dfrac{{4\pi }}{3},f\left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{{4\pi }}{3} + \sin 2\left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{{4\pi }}{3} + \dfrac{{\sqrt 3 }}{2} \\
  x = \dfrac{{5\pi }}{3},f\left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{{5\pi }}{3} + \sin 2\left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{{5\pi }}{3} - \dfrac{{\sqrt 3 }}{2} \\
  x = 2\pi ,f\left( {2\pi } \right) = 2\pi + \sin 2\left( {2\pi } \right) = 2\pi + 0 = 2\pi \\
  $
Out of all the values we got of the function $ f\left( x \right) = x + \sin 2x $ , 0 is the minimum value at x is equal to 0 and $ 2\pi $ is the maximum value at x is equal to $ 2\pi $

Note: The differentiation of sine function will be a positive cosine function but the differentiation of a cosine function will be a negative sine function. Be careful with the signs of the functions. The value of a trigonometric function repeats after a full circle or $ 2\pi $ radians.