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.