## tmagloire1 one year ago Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]

1. IrishBoy123

2. misty1212

HI!!

3. misty1212

did you find $f'(x)$?

4. tmagloire1

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

5. misty1212

hmm no

6. misty1212

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

7. tmagloire1

Yep

8. misty1212

what is the derivative of $$2x$$?

9. tmagloire1

2

10. misty1212

ok good

11. misty1212

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

12. tmagloire1

2cos2x

13. misty1212

ok

14. misty1212

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

15. misty1212

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

16. misty1212

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

17. misty1212

you good from there?

18. tmagloire1

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

19. misty1212

$\color\magenta\heartsuit$

20. tmagloire1

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

21. zepdrix

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

22. zepdrix

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

23. tmagloire1

Ohh ok thank you!

24. zepdrix

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

25. tmagloire1

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

26. zepdrix

divide... by 2 0_o

27. tmagloire1

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

28. tmagloire1

Wow how could i not see that. gr8 moments

29. zepdrix

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

30. zepdrix

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.

31. zepdrix

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

32. zepdrix

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

33. tmagloire1

So pi/2 and 5pi/2

34. zepdrix

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 :)

35. tmagloire1

k=1 will give us 3pi/2

36. tmagloire1

@zepdrix

37. zepdrix

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

38. zepdrix

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!

39. zepdrix

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.

40. tmagloire1

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

41. zepdrix

yay team \c:/