Question

Please help me with this calculus problem!



Find the points of inflection and the local maximum and minimum values of the function y = 2sin(x) + sin^2(x) in the domain of (0, 2pi)

Heres the wolframalpha link: http://www.wolframalpha.com/input/?i=y%3D2sinx%2Bsin%5E2x

Answer 1

3 and -1

Answer 2

How did you get that?

Answer 3

to find inflection points,\[\frac{d^{2}y}{dx^{2}}=0\]

Answer 4

So, you find the second derivative of the base function to get the points of inflection right?

Answer 5

there it is clearly seen on the wolframalpha

Answer 6

Yes, but I dont know how to get that answer...

Answer 7

Yes - the second derivative being 0 at that point is necessary.. but in general, it is not sufficient. One also needs the lowest-order non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection.
Example: \[y = x^4\]

Answer 8

I found the second derivative to be: \[y=2(-\sin(x)\sin(x)+\cos ^{2}(x))-2\sin(x)\]

Answer 9

I don't know how to equate that to zero though, if you guys could help me there.

Answer 10

f'(x) = 2cos(x) + sin(x)cos(x)

use the product rule for the 2nd term then

f"(x) = -2sin(x) - sin^2(x) +cos^2(x)

stationary points

f'(x) = 0 2cos(x) + sin(x)cos(x) = 0 when cos(x) = 0

and sin(x)cos(x) = -2 cos(x)

sin(x) = -2 undefinded

test in 2nd derivative

pts of inflexion

f"(x) = 0 -2sin(x) - sin^2(x) +cos^2(x) = 0 use sin^2 + cos^2 = 1

-2sin(x) - sin^2(x) +( 1 - sin^2(x)) = 0

-2 sin(x) + 1 = 0

sin(x) = -1/2

test with side for change in concavity

Answer 11

Is my derivation wrong?

Answer 12

dont worry your derivation is right

Answer 13

nope - your derivation is correct.

Answer 14

you can simplify\[\cos^{2}x-\sin^{2}x=\cos(2x)\]

Answer 15

How do I equate the second derivative to zero, like how do I solve for x?

Answer 16

its like solving a trigonometric equation

Answer 17

Can you please show me, I am not very good at this...

Answer 18

hmm.. I don't know.. I'd first write it like \[\cos^2(x) = \sin^2(x) + \sin(x)\] 'cause it looks nicer (hope, that's correct). And then I'd get me a nice trig-identity and voilà.. :P

I'd rly like to help, but it seems, i'm too stupid. I'll post, when/if I figure it out, ok? Sry.. :-/

Answer 19

Ok, thanks anyway :/

Answer 20

y = 2sin(x) + sin^2(x)

y' = 2cos(x) + 2sin(x)cos(x)
= 2cos(x) + sin(2x)

y'' = -2sin(x) + 2cos(2x)

Answer 21

So, is the derivation I got wrong?

Answer 22

y''= 2(−sin(x)sin(x)+cos2(x))−2sin(x))

this thing?

Answer 23

Yes

Answer 24

hard to say, can you type up the path that led you?

Answer 25

http://www.wolframalpha.com/input/?i=Differentiate+2cos%28x%29%2B2sin%28x%29cos%28x%29

Answer 26

so your telling me that you opted for the wolfram route :/

Answer 27

:P

Answer 28

it would be better to use the wolf as a chk; but to attempt the derision on your own first.

Answer 29

im sure the wolf is right tho

Answer 30

I know, its just that whenever I try to do it myself, I dont get the same answer as in wolfram, and its not cuz im not right, its cuz theres many different ways to writing the answer, so instead I just use their method

Answer 31

Okay, is it possible for you to show e how to equate my derivation to zero, because I dont know how to do it...

Answer 32

@amistre64 is \[\cos^2(x) - \sin^2(x) = \cos(2x)\] correct?

Answer 33

that is correct

Answer 34

okay :D

I never can remember those, when I need them :-/

Answer 35

-2sin(x) + 2cos(2x) = 0

-sin(x) + cos(2x) = 0

cos(2x) = sin(x)

cos^2(x)-sin^2(x) = sin(x)

1-2sin^2(x) = sin(x)

this is turning into a quadratic

Answer 36

2sin^2(x) + sin(x) - 1 = 0
2 u^2 + u -1 = 0

solve for u and equation u to sin(x)

Answer 37

Can you solve this y''= 2(−sin(x)sin(x)+cos2(x))−2sin(x))=0?

Answer 38

i dont see why i would want to

Answer 39

I need to know how to get the inflection point, and I cant find it without first equating that to zero, and I dont know how to do it :(

Answer 40

most likely by expanding it, and using the same trig identities i did

Answer 41

@JoBo ... it's the exact same equation, cause it simplifies to \[\cos^2(x) = \sin^2(x) + \sin(x)\] like I pointed out above (with the identity I asked @amistre64, you can get his answer in 2 lines).

Just read the post again and thank him for his nice solution ;)

Answer 42

This y = 2sin(x) + sin^2(x)

y' = 2cos(x) + 2sin(x)cos(x)
= 2cos(x) + sin(2x)

y'' = -2sin(x) + 2cos(2x)

Answer 43

This is correct, yes.. as is your wolfram-alpha solution..

\[-2\sin(x) + 2\cos(2x) = 2(-\sin(x)\sin(x)+\cos^2(x))-2(\sin(x))\]
\[-\sin(x) + \cos(2x) = (\cos^2(x) -\sin^2(x))-\sin(x)\]
\[-\sin(x) + \cos(2x) = \cos(2x)-\sin(x)\]

... see??

Answer 44

Ok thanks alot for that explanation! Now, can you help me find the point of inflection?

Answer 45

turn it into a quadratic equation like amistre64 showed you and solve it.. then look at your interval...

Answer 46

Ok, so I should use amistre64's second derivative and do this?

2sin^2(x) + sin(x) - 1 = 0
2 u^2 + u -1 = 0

solve for u and equation u to sin(x)

Answer 47

yes.. use your/his 2nd derivative and solve it like this.. what do you get for x ?

Answer 48

Ok, let me do it

Answer 49

Im confused at:
2sin^2(x) + sin(x) - 1 = 0
2 u^2 + u -1 = 0

solve for u and equation u to sin(x)

I dont quite know how to do that

Answer 50

well, you substituted u for sin(x) and thus got this last equation. It's a simple quadratic equation.. you should be able to solve it.
When you are done solving it (you're gonna get 2 solutions), you just re-substitute again...

Answer 51

Oh, so its \[2u ^{2}+u-1\]?

Answer 52

\[2u^2 +u-1 = 0\] is the last equation, yes..

Answer 53

I got x=-1 and x=0.5

Answer 54

Now what?

Answer 55

Now you substitute back again... so u becomes sin(x) again... thus leaving u with:

\[\sin(x) = -1\]\[\sin(x) = 0.5\]

Answer 56

Oh so instead of x=-1 its sin(x)=-1?

Answer 57

Well, it never was x=-1... it was u=-1 (u being the substitute for sin(x)).

Sry, I should've corrected that in you reply.. the substitution is just temporary, so we get a nice, simple quadratic equation.

Answer 58

Can I write:\[2\sin ^{2}(x)+\sin(x)-1=0\]
\[\sin(x)=-1 and \sin(x)=0.5\]?

Answer 59

well, yes... that is the result - now all we have to do, is to solve for x

Answer 60

How do we do that?

Answer 61

Hmm... I think you do know the inverse function for sin(x), don't you? ;)

Answer 62

Nope :P

Answer 63

you don't? rly? hmm...

well, I don't know what to tell you xD
Every trigonometric function has an inverse function, like:\[\sin^-1 (\sin(x)) = x\]
This function sin^(-1) is also called arcsin()..

So:\[\sin(x) = -1\]\[x = \arcsin(-1) = - \frac{\pi}{2} = -90°\]

Answer 64

Ok

Answer 65

-> http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

But we got a problem.. arcsin() only yields values between -90° and +90°, while you should search the interval between 0° and 360°... so, we are missing an x-intercept at 150°..

Can anyone tell me, where I lost that information in the process? @amistre64 maybe? :)

Answer 66

Remember, We're trying to find the point of inflection

Answer 67

Lets continue this tomorrow, I'm really sleepy.

Answer 68

Goodnight :)

Answer 69

Ok now see - I lied to you xD

Given\[\sin(x) = y\]\[y \neq \arcsin(x)\] but rather \[y = \arcsin(x) + 2k \pi \]\[y = \pi - \arcsin(x) + 2k \pi\]

So that's why we get 30° and 150° from one of the equations nad 270° from the other...

Answer 70

Well, usually it doesn't take that long, but I haven't done this since like.. forever :P

n8y n8

Answer 71

without quadratic equation\[y''=-2\sin x+2\cos \left( 2x \right)\]\[y''=0\]\[-2\sin x+2\cos \left( 2x \right)=0\]\[\sin x=\cos \left( 2x \right)\]\[\sin x=\sin \left( \frac{\pi}{2} -2x\right)\]\[x=\frac{\pi}{2}-2x+k2\pi\]\[x=\frac{\pi}{6}+k\frac{2\pi}{3}\]\[k=0\rightarrow x=\frac{\pi}{6}\]\[k=1\rightarrow x=\frac{5\pi}{6}\]\[k=2 \rightarrow x = \frac{3\pi}{2}\]

Answer 72

awesome - thanks exraven :)