Answer:
The x-coordinate is \dfrac{\pi}{6}[/tex].
Step-by-step explanation:
We are given a function f(x) as:
[tex]f(x)=2x+\sin (4x)[/tex]
Now on differentiating both side with respect to x we get that:
[tex]f'(x)=2+4 \cos (4x)[/tex]
When [tex]f'(x)=0[/tex]
this means that [tex]2+4\cos (4x)=0\\\\4\cos (4x)=-2\\\\\cos(4x)=\dfrac{-1}{2}[/tex]
Hence, cosine function takes the negative value in second and third quadrant but we have to only find the value in the interval [tex][0,\pi][/tex].
also we know that [tex]\cos (\dfrac{2\pi}{3})=\dfrac{-1}{2}[/tex]----(1) (which lie in the second quadrant)
so on comparing our equation with equation (1) we obtain:
[tex]4x=\dfrac{2\pi}{3}\\\\x=\dfrac{\pi}{6}[/tex]
Hence, the x-coordinates where [tex]f'(x)=0[/tex] for [tex]f(x)=2x+\sin(4x)[/tex] is [tex]\dfrac{\pi}{6}[/tex].
