The solutions of the equation [tex]f\left( x \right) = 2x + \sin \left( {2x} \right)[/tex] in the interval [tex]\left[ {0,2\pi } \right][/tex] are [tex]\boxed{\left( {\frac{\pi }{2},\pi } \right)}[/tex] and [tex]\boxed{\left( {\frac{{3\pi }}{2},3\pi } \right)}.[/tex]
Further explanation:
Given:
The function is [tex]f\left( x \right) = 2x + \sin \left( {2x} \right).[/tex]
The first derivative is zero.
Explanation:
The given function is [tex]f\left( x \right) = 2x + \sin \left( {2x} \right).[/tex]
Differentiate the function with respect to [tex]x[/tex].
[tex]\begin{aligned}f'\left( x \right) &= 2 + 2\cos \left( {2x} \right)\\&= 2\left( {1 + \cos 2x} \right)\\\end{aligned}[/tex]
Substitute [tex]0[/tex] for [tex]f'\left( x \right).[/tex]
[tex]\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}[/tex]
In the interval [tex]\left[ {0,2\pi } \right][/tex] the x-coordinates are [tex]\boxed{\frac{\pi }{2}}{\text{ and }}\boxed{\frac{{3\pi }}{2}}.[/tex]
The solutions of the equation [tex]f\left( x \right) = 2x + \sin \left( {2x} \right)[/tex] in the interval [tex]\left[ {0,2\pi } \right][/tex] are [tex]\boxed{\left( {\frac{\pi }{2},\pi } \right)}[/tex] and [tex]\boxed{\left( {\frac{{3\pi }}{2},3\pi } \right)}.[/tex]
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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.
