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

Question

irishboy123 · Answer

go ahead and differentiate

misty1212 · Answer

HI!!

misty1212 · Answer

did you find $$f'(x)$$?

tmagloire1 · Answer

f'(x) =4cos^2(x)

misty1212 · Answer

hmm no

misty1212 · Answer

$$ f(x) = 2x + \sin(2x) $$ right?

tmagloire1 · Answer

Yep

misty1212 · Answer

what is the derivative of $2x$?

tmagloire1 · Answer

2

misty1212 · Answer

ok good

misty1212 · Answer

now how about the derivative of $\sin(2x)$?

tmagloire1 · Answer

2cos2x

misty1212 · Answer

ok

misty1212 · Answer

so the derivative is not $4\cos^2(x)$ is it?

misty1212 · Answer

it is $$f'(x)=2+2\cos(2x)$$

misty1212 · Answer

final job is to set 
$$2+2\cos(2x)=0$$ and solve for $x$

misty1212 · Answer

you good from there?

tmagloire1 · Answer

Yes thank you! I was just getting the wrong derivative then my bad!

misty1212 · Answer

$$\color\magenta\heartsuit$$

tmagloire1 · Answer

actually can someone help me find 2+2cos(2x)=0

zepdrix · Answer

`subtract`
2cos(2x)=-2
`divide 2`
cos(2x)=-1

zepdrix · Answer

cosine of angle is -1 when the angle isssss...... pi, ya?

tmagloire1 · Answer

Ohh ok thank you!

zepdrix · Answer

careful though!
the angle is (2x). that leads to 2x=pi
still need to go further to solve for x.

tmagloire1 · Answer

im still confused though. if it is 2x=pi than how do you get the x coordinates for the original problem @zepdrix

zepdrix · Answer

divide... by 2 0_o

tmagloire1 · Answer

oh.my.god.im retarded lolol. so x=pi/2

tmagloire1 · Answer

Wow how could i not see that.  gr8 moments

zepdrix · Answer

Oh oh we should be careful though...$$\large\rm \cos(2x)=-1\qquad\to\qquad 2x=\pi+2k \pi,\qquad k\in \mathbb Z$$

zepdrix · Answer

It's not just pi,
we can also add any multiple of 2pi to that angle, ya?
It's like spinning an extra time around the circle and landing in the same location.

zepdrix · Answer

This will be important because we might get several solutions within this interval.

zepdrix · Answer

Dividing by 2,$$\large\rm x=\frac{\pi}{2}+k \pi$$

tmagloire1 · Answer

So pi/2 and 5pi/2

zepdrix · Answer

we're using integers for k.
So k=0 gives us pi/2, looks good!
k=1 gives us ... ?
k=2 gives us 5pi/2 <- this is larger than 2pi, out of our interval, so we can throw it away :)

tmagloire1 · Answer

k=1 will give us 3pi/2

tmagloire1 · Answer

@zepdrix

zepdrix · Answer

Good! :) Here is a graph to help you see what is going on. https://www.desmos.com/calculator/xm7sfacgsm

zepdrix · Answer

I graphed the function 2x+sin(2x) from 0 to 2pi.
You can clearly see that at pi/2 and 3pi/2 we have critical points!

zepdrix · Answer

That's what f'(x)=0 gives us, critical points.
locations where the slope of the function is zero.
graphically it looks like the top of a hill, or the bottom of a valley.

tmagloire1 · Answer

Ohh okay so critical points is a another way of looking at it! Thank you so much!

zepdrix · Answer

yay team \c:/