The solutions of the equation \(f\left( x \right) = 2x + \sin \left( {2x} \right)\) in the interval \(\left[ {0,2\pi } \right]\) are \(\boxed{\left( {\frac{\pi }{2},\pi } \right)}\) and \(\boxed{\left( {\frac{{3\pi }}{2},3\pi } \right)}.\)
Further explanation:
Given:
The function is \(f\left( x \right) = 2x + \sin \left( {2x} \right).\)
The first derivative is zero.
Explanation:
The given function is \(f\left( x \right) = 2x + \sin \left( {2x} \right).\)
Differentiate the function with respect to \(x\).
\(\begin{aligned}f'\left( x \right) &= 2 + 2\cos \left( {2x} \right)\\&= 2\left( {1 + \cos 2x} \right)\\\end{aligned}\)
Substitute \(0\) for \(f'\left( x \right).\)
\(\begin{aligned}2\left( {1 + \cos 2x} \right) &= 0 \\1 + \cos 2x &= 0\\\cos 2x &= - 1\\2x &= {\cos ^{ - 1}}\left( { - 1} \right)\\2x &= \frac{{\left( {2n - 1} \right)\pi }}{2} \\\end{aligned}\)
In the interval \(\left[ {0,2\pi } \right]\) the x-coordinates are \(\boxed{\frac{\pi }{2}}{\text{ and }}\boxed{\frac{{3\pi }}{2}}.\)
The solutions of the equation \(f\left( x \right) = 2x + \sin \left( {2x} \right)\) in the interval \(\left[ {0,2\pi } \right]\) are \(\boxed{\left( {\frac{\pi }{2},\pi } \right)}\) and \(\boxed{\left( {\frac{{3\pi }}{2},3\pi } \right)}.\)
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Application of derivatives
Keywords: derivative, x – coordinates, interval, far, 2x, sin2x, coordinates, 0, 2pi, y-coordinate.
