Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]. so far I found f'(x)=2cos(2x)+2 cos(2x)

Question

Alex17521 [72]

4 years ago

7

Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]. so far I found f'(x)=2cos(2x)+2 cos(2x)

=-1

Mathematics

2 answers:

rusak2 [61] · Answer 1 · 2021-12-02T23:19:14+03:00

From there, you simply need algebra and a calculator that works in radians.

Take the inverse cos of both sides to get 2x = arccos(-1)

Then divide both sides by 2 to get x = arccos(-1) / 2

Put that into a calculator and you get π/2. But because your bounds are 0 to 2π, you have to add π your solution to get the solution on the other side of the unit circle, which would be (3π/2).

Now that you have the x value, put (π/2) and (3π/2) into f(x) to get the y coordinate.

f(π/2) = 2(π/2) + sin(2(π/2) = π, which means this solution is just (π/2, π)
f(3π/2 = 2(3π/2) + sin(2(3π/2) = 3π, which means this solution is (3π/2, 3π)

Katarina [22] · Answer 2 · 2021-12-03T01:31:40+03:00

The solutions of the equation $f\left( x \right) = 2x + \sin \left( {2x} \right)$ in the interval $\left[ {0,2\pi } \right]$ are $\boxed{\left( {\frac{\pi }{2},\pi } \right)}$ and $\boxed{\left( {\frac{{3\pi }}{2},3\pi } \right)}.$

Further explanation:

Given:

The function is $f\left( x \right) = 2x + \sin \left( {2x} \right).$

The first derivative is zero.

Explanation:

The given function is $f\left( x \right) = 2x + \sin \left( {2x} \right).$

Differentiate the function with respect to $x$ .

$\begin{aligned}f'\left( x \right) &= 2 + 2\cos \left( {2x} \right)\\&= 2\left( {1 + \cos 2x} \right)\\\end{aligned}$

Substitute $0$ for $f'\left( x \right).$

$\begin{aligned}2\left( {1 + \cos 2x} \right) &= 0 \\1 + \cos 2x &= 0\\\cos 2x &= - 1\\2x &= {\cos ^{ - 1}}\left( { - 1} \right)\\2x &= \frac{{\left( {2n - 1} \right)\pi }}{2} \\\end{aligned}$

In the interval $\left[ {0,2\pi } \right]$ the x-coordinates are $\boxed{\frac{\pi }{2}}{\text{ and }}\boxed{\frac{{3\pi }}{2}}.$

The solutions of the equation $f\left( x \right) = 2x + \sin \left( {2x} \right)$ in the interval $\left[ {0,2\pi } \right]$ are $\boxed{\left( {\frac{\pi }{2},\pi } \right)}$ and $\boxed{\left( {\frac{{3\pi }}{2},3\pi } \right)}.$

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