Question

Find the x-coordinates where f ' (x)=0 for f(x)=2x+sin(4x) in the interval [0, pi]

Answer 1

The x-coordinate is \dfrac{\pi}{6}[/tex].

Explanation:

We are given a function f(x) as:

`f(x)=2x+\sin (4x)`

Now on differentiating both side with respect to x we get that:

`f'(x)=2+4 \cos (4x)`

When
`f'(x)=0`

this means that
`2+4\cos (4x)=0\\\\4\cos (4x)=-2\\\\\cos(4x)=(-1)/(2)`

Hence, cosine function takes the negative value in second and third quadrant but we have to only find the value in the interval
`[0,\pi]`.

also we know that
`\cos ((2\pi)/(3))=(-1)/(2)`----(1) (which lie in the second quadrant)

so on comparing our equation with equation (1) we obtain:

`4x=(2\pi)/(3)\\\\x=(\pi)/(6)`

Hence, the x-coordinates where
`f'(x)=0` for
`f(x)=2x+\sin(4x)` is
`(\pi)/(6)`.