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lutik1710 [3]
3 years ago
10

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

Mathematics
1 answer:
MAVERICK [17]3 years ago
5 0

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}.

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