Question

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

Answer 1

Answer:

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

Step-by-step 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 $2+4\cos (4x)=0\\\\4\cos (4x)=-2\\\\\cos(4x)=\dfrac{-1}{2}$

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 $\cos (\dfrac{2\pi}{3})=\dfrac{-1}{2}$----(1) (which lie in the second quadrant)

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

$4x=\dfrac{2\pi}{3}\\\\x=\dfrac{\pi}{6}$

Hence, the x-coordinates where $f'(x)=0$ for $f(x)=2x+\sin(4x)$ is $\dfrac{\pi}{6}$.