Question

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

This is what I have so far:
y=2sin(x)+sin^2(x)
y'=2cos(x)+2sin(x)cos(x)
y'=2cos(x)+sin(2x)
y''=-2sin(x)+2cos(2x)
-2sin(x)+2cos(2x)=0
-sin(x)+cos(2x)=0
cos(2x)=sin(x)
1-sin^2(x)=sin(x)
2sin^2(x)+sin(x)-1=0
sin(x)=-1 & sin(x)=1/2
Please help!

Answer 1

Find the zeros of the second derivative. On a trig function like this one, those will be the points of inflection of the function. The zero of the second derivative represents the places where the curvature is going from convex upward to convex downward, or visa-versa.

Answer 2

Yes, so I got that I think, is it right?

Answer 3

I follow it until the step where cos (2x) becomes 1-sin^2(x). I don't think this is justified.

Answer 4

So what should I do?

Answer 5

cos2x = 1-2sin^2x ...

Answer 6

Well, one solution is x= pi/6 in that case \[\cos(\frac{\pi}{3})=\sin(\frac{\pi}{6})\]

Answer 7

Which I think is true.

Answer 8

So if I corrected it to cos2x = 1-2sin^2x
what would my final answer be?

Answer 9

In your workings :
cos(2x)=sin(x)
1-sin^2(x)=sin(x) <- incorrect
2sin^2(x)+sin(x)-1=0 <- correct
sin(x)=-1 & sin(x)=1/2 <- correct

Answer 10

Oh, so its a typo

Answer 11

I haven't taught trig in quite a while.... what is the identity that justifies this? I'm sorry to be intransigent, but I'm not seeing it.

Answer 12

To find the inflection point you said we need to equate the equation to 0, ive done that and got sin(x)= -1, 1/2, are those the inflection points?

Answer 13

@AnimalAin it's cos2x = cos^2x - sin2x = 1-2sin^2x = 2cos^2x-1

Answer 14

sin x = 1/2 corresponds to the answer I got earlier. I don't think arcsin -1 meets the specifications of the earlier condition (that would be cos 2x = sin x.

Thanks Callisto; I always need to look up those double angle and half angle formulas; don't have my books handy here.

Answer 15

The other postition of x that generates sin x = 1/2 is x= 5pi/6. Twice that position is in the fourth quadrant, and would give a negative value for cos 2x, so it couldn't meet the specification of the problem, either.

Answer 16

Not yet
sin(x)= -1 or sinx = 1/2
x= 3pi /2 x = pi/6 or x=5pi /6
| x<pi/6 | pi/6 < x<5pi/6 | 5pi/6<x<3pi/2 | x>3pi/2
y''| + | - | + | +

Therefore , inflection points are at x=pi/6, x=5pi/6
y=2sin(x)+sin^2(x) -(1)
Put x = pi/6 into (1)
y= 1.25
Put x=5pi/6 into (1)
y= 1.25

So inflection points are (pi/6 , 1.25) and (5pi/6 , 1.25)

Answer 17

AnimalAin You're welcome

Answer 18

How did you get the x= values?

Answer 19

sin 270degrees = -1
sin 30 degrees = 1/2
sin 150 degrees = 1/2
Use sin^(-1) 1/2 and sin^(-1) -1

Answer 20

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

Answer 21

I see

Answer 22

So how do we find the local max and mins?

Answer 23

Put y'=0 and solve x, then find the respective y value(s)

Answer 24

I dont understand...

Answer 25

Which part?

Answer 26

@JoBo

Answer 27

I'm sorry, I was confused at this end..... arcsin (-1)= 3pi/2 and cos(3pi)=-1

Answer 28

put y'=0

Answer 29

@AnimalAin arcsin (-1)= 3pi/2 <- this is correct, but this is not the inflection point as the y'' is doesn't change, see the table i typed may help you understand, i hope.
And I don't think I have mentioned ' cos(3pi)=-1' ... Sorry :(

Answer 30

@JoBo
For a maximum/minimum point, the slope at that point =0
Therefore y'=0, that is
2cos(x)+sin(2x) =0
Solve x from here. Understand?

Answer 31

Oh, thank you so much I finally understand! :) lol but I dont know how to solve for x...

Answer 32

@callisto?

Answer 33

Sorry... was away for a while...

Answer 34

Thats ok :)

Answer 35

Can you teach me how to solve for x? for 2cos(x)+sin(2x) =0

Answer 36

2cos(x)+sin(2x) =0
2cos(x)+2sin(x)2cos(x) =0
2cosx (1+sin(x))=0
2cosx =0 or 1+sinx =0
cosx=0 or sinx =-1
x = pi/2 or x= 3pi/2 or x = 3pi/2 (repeated)
x<pi/2 | pi/2 <x < 3pi/2 | x >3pi/2
y' | + | - | +
(slope) / \ /
So max is at x=pi/2 min is at 3pi/2
Work out the y values now

Answer 37

@JoBo

Answer 38

Sorry about that

Answer 39

Working out the y values will get me the coordinates for the max and min?

Answer 40

yes, put x value(s) into 2sin(x)+sin^2(x) to get the corresponding y-value(s)

Answer 41

I still dont understand how you did this part...
y=2sin(x)+sin^2(x) -(1)
Put x = pi/6 into (1)
y= 1.25
Put x=5pi/6 into (1)
y= 1.25

Answer 42

You know I solve the x values for inflection point right?

Answer 43

Yes

Answer 44

I dont understand how you got 1.25

Answer 45

you know the curve is a graph of y=2sin(x)+sin^2(x) , right?

Answer 46

Yes

Answer 47

now you have x value, how would you get the y value?

Answer 48

The x value being ie. pi/6 where do I plug that in?

Answer 49

when you see x, replace it by pi/6 in the function of the graph

Answer 50

How would I input that into my calculator?

Answer 51

is it in radian or in degree mode?

Answer 52

I can change it

Answer 53

but yeah, degree mode

Answer 54

Sorry
pi/6 = 30 degree
Put
2sin(30)+sin^2(30) in your calculator, then it will give you the value 1.5

Answer 55

I cant do sin^2(30) it does sin(30)^2

Answer 56

@callisto?

Answer 57

Sorry.. I didn't notice your response.. sorry
sin^2(30) = sin(30)^2

Answer 58

When I do 2sin(30)+sin(30)^2 I get 1

Answer 59

Oopss.... it should be 1.25!! How silly me!!1

Answer 60

Yea, but I get 1

Answer 61

try again~~~
2sin(30)+sin(30)^2 = 2(1/2) + (1/2)^2 = 1+0.25 = 1.25

Answer 62

Oh yea youre right

Answer 63

Got it?!

Answer 64

For this post:
2cos(x)+sin(2x) =0
2cos(x)+2sin(x)2cos(x) =0
2cosx (1+sin(x))=0
2cosx =0 or 1+sinx =0
cosx=0 or sinx =-1
x = pi/2 or x= 3pi/2 or x = 3pi/2 (repeated)
x<pi/2 | pi/2 <x < 3pi/2 | x >3pi/2
y' | + | - | +
(slope) / \ /
So max is at x=pi/2 min is at 3pi/2
Work out the y values now

I put the x values back into which equation?

Answer 65

Original one. Whenever you need to find the corresponding y-value of x in the graph, put x into the original one , that is not y' nor y''

Answer 66

What does finding the y values do?

Answer 67

So I do 2cos(90)+sin(180)?

Answer 68

@Callisto?

Answer 69

nope :S then x value input should be the same every time :S

Answer 70

Im confused...

Answer 71

are you talking about sin(2x)? I just multiplied it

Answer 72

I see... but the original equation is in sin only... no cos as i see...

Answer 73

I did 2cos(x)+sin(2x) =0

Answer 74

Are you talking about 2sin(x)+sin^2(x)
?

Answer 75

yes.... this is the original one :S

Answer 76

I get 0 for the first one and -1 for the second one

Answer 77

What do I do with these numbers?

Answer 78

this is the corresponding values of x
the coord = (pi/2, 0) for max
(3pi/2 , -1) for min
Assume you've not made any mistakes

Answer 79

I see, THANK YOU SO MUCH Callisto!!! <333

Answer 80

welcome!!!