Question

I have a calculus problem I need help with! Can anyone help me?

Answer 1

Is anyone there?

Answer 2

Post your question :) Someone will come along to help!!

Answer 3

I'm taking the first and second derivate for 2sin(x)+sin(2x) on [0,2pi]
I think I have my first derivate which is
2cos(x)+(cos(2x)*2)= 2cosx+2cos(2x)=0
Divide each 2 out u get
=cosx+cos(2x)=0
=cos+2cos^2x-1=0
=2cos^2x+cosx-1=0
Then I did the quadratic formula which gave me x=-1 x=1/2
So {pi, pi/3, 5pi/3}
So now I need the second derivate

Answer 4

Hmm ok cool everything looks good so far! :)

Answer 5

\[\Large\bf f(x)\quad=\quad 2\sin x+\sin 2x\]\[\Large\bf f'(x)\quad=\quad 2\cos x+2\cos 2x\]\[\Large\bf f''(x)\quad=\quad ?\]So we're having a little trouble with the second derivative?
Remember derivative of cosine or no? :o

Answer 6

-sin

Answer 7

I think my second derivate is
-2sinx-4sin(2x)=0

Answer 8

\[\Large\bf f''(x)\quad=\quad -2\sin x-4\sin 2x\]Mmm ok looks good!

Answer 9

So you found critical points using the first derivative.
Are we looking for inflection points with the second derivative here?

Answer 10

So we can determine intervals of concavity and all that jazz?

Answer 11

I'm lost after I find the second derivative. Once I get to -2sin(x)-4sin(2x)=0 I'm stuck

Answer 12

Can I divide 2 out and get
-sin-2sin(2x)=0

Answer 13

Yes, let's go a little further and divide a -2 out actually, so we have some nice pretty addition.

From there, we want to recall our `Sine Double Angle Formula`:
\[\Large\bf \color{royalblue}{\sin2x\quad=\quad 2\sin x \cos x}\]

Answer 14

Ok so we will have
Sinx+2(2sinxcosx)=0

Answer 15

So maybe 2sinx+2sinxcosx=0

Answer 16

Hmm woops that 2 moved to the wrong spot didn't it?

\[\Large\bf \sin x+2(2\sin x \cos x)\quad=\quad 0\]\[\Large\bf \sin x+4\sin x \cos x\quad=\quad 0\]

Answer 17

Oh ok!!

Answer 18

Hmm do the terms have anything in common that we could factor out of them?\[\Large\bf \color{#DD4747 }{\sin x}+4\color{#DD4747 }{\sin x} \cos x\quad=\quad 0\]

Answer 19

So do we add the sins together now?

Answer 20

Hmm we can't add them, they're not like terms because of the cosine.

Answer 21

I'm sorry yes

Answer 22

So we factor out the sins and that leaves us with
Sinx(4cosx)=0

Answer 23

Woops!
We should be left with `2 terms` inside the brackets.
We started with 2 terms, we should end with 2 terms after factoring.

The first term, you factored a sinx out of sinx, what does that leave you with?
Hint: Not zero.

Answer 24

Sinx(1+4cosx)??? I'm lost if that's not it

Answer 25

Ok good good good :)

Answer 26

I'm working it out Will give u my answer in a sec

Answer 27

So Sinx=0
Divide sin out x=0

Then 1+4cosx=0
-1. -1
4cosx= -1
Divide 4 out
Cosx= -1/4

Answer 28

Divide sin out? +_+

Answer 29

no no no silly!

Answer 30

\[\Large\bf \sin x\quad=\quad0\]Sine giving us zero corresponds to a few special angles.
Yes, \(\Large \bf x=0\) is one of those.
There should be a couple more in our interval though.

Answer 31

Ok I'm lost b/c I set the equation equal to 0 so that's why I got that but if it's not it how do I get the answers

Answer 32

Sinx=o

Answer 33

1+4cosx=0

Answer 34

Wouldn't that be -1/4??

Answer 35

Yes you set it equal to zero, good.
But sinx=0 should give us more solutions than just x=0.\[\Large\bf \sin x=0 \qquad\to\qquad x=0,\quad \pi,\quad 2\pi,\quad 3\pi,\quad 4\pi,\quad...\]

Answer 36

The sine of pi is also zero right?
How many of those angles fall into our given interval?

Answer 37

It's just to 2pi we only have to have [0,2pi]

Answer 38

Does it look like that, or is it a rounded bracket on the 2pi?

[0,2pi)

Answer 39

No it's in [0,2pi]

Answer 40

The original function is
2sin(x) + sin(2x) on [0,2pi]

Answer 41

So how many solutions would the sinx be giving us?
I guess it would give us:\[\Large \bf x=0,\quad \pi,\quad 2\pi\]Yes?

Answer 42

Yes!! So now do we set the 1+4cosx=0 and solve?

Answer 43

Yes, that one is a little weird unfortunately :(
since:\[\Large \bf \cos x=-\frac{1}{4}\]Does not correspond to a special angle.

Answer 44

I guess we could maybe write it like this:\[\Large\bf x=\arccos\left(-\frac{1}{4}\right)\]

Answer 45

So what would I do??

Answer 46

We haven't learned that?

Answer 47

Mmm you should have learned about inverse trig functions back in trigonometry D':
We can approximate the solutions that the cosine is giving us by putting that into a calculator.
That doesn't seem like a bad idea.

Answer 48

Well my friend here says we did but I cannot remember!! Lol sorry it's late here. So inverse trig. Function how do we do that again?

Answer 49

Or u said put in calculator??

Answer 50

Yah go back to the ole handy calculator. c:
There should be a \(\Large \cos^{-1}\) button somewhere.

Answer 51

Ok I see it

Answer 52

So cos^-1(-1/4) in the calculator??

Answer 53

ya

Answer 54

1.3181...

Answer 55

Woops put your calculator in degree mode.

Answer 56

104.47751...

Answer 57

Ok good. So one of our inflection points is located around 105 degrees.

There is a small problem though.
There should be `two` angles within one full rotation of 2pi that give us this measure.
But the calculator is only spitting out the one in the 2nd quadrant.
There is another in the 3rd quadrant.

Answer 58

And how do we find that angle... ummm

Answer 59

Lol I'm looking but that's tricky

Answer 60

210degrees???

Answer 61

So I guess to get the other one we would do:\[\Large\bf x\quad=\quad 360^o-\cos^{-1}\left(-\frac{1}{4}\right)\]

Answer 62

210? Hmm I think that's close.
I think it's more like 255ish.

Answer 63

Oh ok? How did u know that?? Just b/c it's one full revolution?

Answer 64

I got 284.477... ??

Answer 65

But that's in quad 4

Answer 66

Hmm our first angle was around 104.5 right?
So 360 - 104.5 is approx 255.5, yes? +_+

Answer 67

Oh I put it in the calculator like..
360-cos^-1(-1/4)

Answer 68

So I should subtract the 360 from the 104.5 then take that times (-1/4)

Answer 69

360-cos^-1(-1/4)
should have given you the correct answer.. hmm weird...

Answer 70

Oh I just redid it and got 255.52...

Answer 71

I guess I entered it wrong at first sorry

Answer 72

Ok good so there are the inflection points.\[\Large x=0,\pi,2\pi\]\[\Large x \approx 104.5^o,255.5^o\]

Answer 73

Ok wow!! Ok that's so crazy to me b/c we haven't had any that look like that this semester and this is a take home test so I wonder why he would do that??

Answer 74

Yah that's a weird problem :( Hmm
Mean teacher ? lol

Answer 75

Ok my friend is doubting this. She had someone (a teacher) say something about there was two solutions to the first derivate and one for the second?? Any thoughts??

Answer 76

??

Answer 77

hmm? :o

Answer 78

sec I'll show you the graph.

Answer 79

Ok

Answer 80

https://www.desmos.com/calculator/hbaalbasda

Answer 81

You can click on the curve at different points.
You can see that we have critical points at pi/3, pi, and 5pi/3 as you said.

Answer 82

Yes I see that clearly!! Idk why he would do this but who knows maybe he's bored!!! Lol

Answer 83

Unless the problem was supposed to be,\[\Large f(x)\quad=\quad 2\sin x+\sin^2x\]In which case yes, THAT function has 2 critical points in the given interval, and 1 inflection point I think...

Answer 84

Or the teacher ( or whoever ) thought that's what the problem was hehe

Answer 85

Lol mayb!!

Answer 86

Oh also,
\(\Large \bf\color{royalblue}{\text{Welcome to OpenStudy! :)}}\)

Answer 87

But no it's what I told u
2sin(x)+sin(2x)

Answer 88

Oh thank you I was pulling my hair out

Answer 89

Ok I have one more!! Can u help??

Answer 90

No too late :c need sleep.

Close this thread and open up a new one c:
Someone will probably come along.

Answer 91

Oh ok thank you