Question

Calculus question:
Relative extrema and intervals
Help much appreciated!

Answer 1

find the intervals for which f is increasing or decreasing and locate all relative extrema f(x)= (cosx)/(1 +sin^2x)

Answer 2

the interval is (0,2pi)
right now I am wrestling with differentiating (yikes, long quotient rule!)
i'll keep working, and I appreciate any help!

Answer 3

\[\frac{ d }{ dx }(\frac{ u }{ V }) =\frac{ V(\frac{ DU }{ DX }) - u(\frac{ DV }{ DX }) }{ V ^{2} }\]

Answer 4

ah , thank you! I was afk. give me a minute to look at these.

Answer 5

hm, this is what I had gotten so far:
f'= (1+ sin^2x)(-sinx)- (cosx)(cos^2x)/(1+sin^2x)^2
but I do think I made a mistake.
could you explain what I did wrong?

Answer 6

it is correct.

Answer 7

oh it is? ok so I just need to simplify, set= to 0, solve for x...
i'll get on that, and it would be great to have some help on hand still

Answer 8

... would it be this?
simplified partway
(-sinx -sin^3x -cos^3x)/(1+ sin^2x +sin^2x +sin^4x)
I am not sure about this though

Answer 9

u r correct.

Answer 10

this problem is really killing me with the simplification...
and I am not sure how to solve for this when it =0
praxer, I am correct? O_O wow
ok then
I just have to set the denominator =0 and solve, right?
let me try...

Answer 11

so if I did this...
(1+ 2sin^2x +sin^4x)=0
2sin^2x +sin^4x = -1
... yipes. what do I do now? trigonometric problems always scare me!

Answer 12

anyone? I'm still stuck here. Is there something I should be factoring out or dividing?

Answer 13

inverse...circular...trigonometry? you mean using the Unit Circle?

Answer 14

let me rearrange it once,
f'(x) = \(\dfrac{-sinx(1+sin^2x) - cosx(2sinx cosx)}{(1+sin^2x)^2}\)

Answer 15

check, please. for the second term of numerator.

Answer 16

you have cosx * (1+sin^2x)' = cosx (2 sinx cos x) not as yours

Answer 17

I am waiting for your checking before doing the next step

Answer 18

http://www.wolframalpha.com/input/?i=%281%2BSIN^2x%29%27

Answer 19

ok I am checking mine, I did something incorrectly then?
oh did I differentiate the second term incorrectly? I think I see it
it is -cosx(2sinxcosx) instead of -(cosx)(cos^2x)?

Answer 20

yup.

Answer 21

from there should I set the denominator equal to 0 and solve for x?

Answer 22

so, for numerator : it is -sinx[(1+sin^2x)-2cos^2x] = -sin x (2 sin^2x -cos^2 x)
got me so far?

Answer 23

yes ok

Answer 24

I don't know why you focus on denominator but numerator. My prof taught me consider both.

Answer 25

oh ok.
well I know that the steps for this problem are:
find f', set=0, solve for x values which are critical pts, make and test intervals, and so on...
I was not entirely sure how to set the derivative=0, so I just thought finding where the denominator =0 would be easiest.
how should I do it?

Answer 26

we don't have that option because we need f'(x) =0 and it happen iff the numerator =0

Answer 27

not denominator.

Answer 28

oh ok, silly me!
so 0= -sinx(2sin^2x-cos^2x)
and...i'm sorry but i'm terrible with trigonometric problems. could you please show me what to do next? distribute in the -sinx and put one term on the other side, or something? is that what we do?

Answer 29

I am not sure whether I can go to the end, just try.
sin x =0 iff x =0 or 2pi (2 critical points)
2sin^2x - cos^2x =0 need some step to convert it

Answer 30

ok but thank you so much for your help so far
as long as I can find the critical points, I can easily do the intervals and the rest
I just have a hard time solving with trig. identities and setting=0
so 2sin^2x-cos^2x=0 and I should be doing something to simplify it further?

Answer 31

we have sin^2 x = 1 - cos^2 x , so 2sin^2 x-cos^2 x = 2 -2 cos^2x -cos^2 x = 2- 3 cos^2 x
so far so good

Answer 32

and we want it =0, that means 2 -3cos^2 x =0 or cos^2x = 2/3 so, cos x = \(\pm \sqrt{2/3}\)

Answer 33

lalala.... i don't think it 's hard to you any more.

Answer 34

oh dear this may be a ridiculous question but I am not seeing sqrt(2/3) on my unit circle. I should be looking for this on my unit circle to find x, right?

Answer 35

cannot use calculator?

Answer 36

oh let me try that then -_- silly, i'm silly...

Answer 37

I may make mistake at somewhere, but that is the step. you check it and finish the rest, ok?

Answer 38

For the derivative I am getting
f'(x) = (-sin(x)(1 + sin^2(x) + 2cos^2(x)) / (1 + sin^2(x)) = 0

Answer 39

This is simplifying @Loser66 posting earlier with the statement: "let me rearrange it once..."

Answer 40

The denom above has to be squared...

Answer 41

yes, I know where is my mistake, thanks ranga, we take -sin out, so, the second term must be + , it means what ranga stated.

Answer 42

Setting the numerator to zero we get
-sin(x) * (1 + sin^2(x) + 2cos^2(x)) = 0
The second parenthesis can never be zero.
so only need to solve for sin(x) = 0

Answer 43

about the first set of critical numbers being x= 0 or 2pi
my interval is (0,2pi) which is noninclusive... so wouldn't these two be excluded as points and as test intervals?

Answer 44

so we would only need to solve for the second one?

Answer 45

If it is an open interval then x = pi is the only CP

Answer 46

it is an open interval, yup.

Answer 47

sin(x) = 0 at x = 0, pi, 2pi. Since 0 and 2pi are not in the domain, x = pi is the only critical point.

Answer 48

right, so I think we just did some unrelated work.
so x= pi would be the only c.p. so I will make intervals (-infinity, pi) (pi, infinity) and test those.

Answer 49

hehehe.. sorry.

Answer 50

Your domain is (0, 2pi) and so you just need to test the interval (0, pi] and [pi, 2pi)

Answer 51

right, whoops. so I can pick a pt. in each and test in
(-sinx)(1+sin^2x+2cos^x)/(1+sin^2x)^2 ?

Answer 52

Yes. x = pi/2 may be one easy point to check.

Answer 53

ok I will try it thank you! should I be doing this with calculator or unit circle? does it matter?
also thank you both so much for being so steadfast! this problem has been really confusing for me so I truly appreciate your help!

Answer 54

unit circle. The sine and cosine are either 0 or 1 at x = pi/2 and so unit circle is much easier. You don't even need the exact f'(pi/2). All you need is: is it positive or negative. The bottom is always positive and so you can ignore it.

Answer 55

ok so f'(pi/2)= -2( 1+
and how will I solve when it is 'sin^2' or 'cos^2'?
so I should only substitute into the numerator and simplify that?

Answer 56

sin(pi/2) = 1; cos(pi/2) = 0
numerator of f'(x) = -sin(x) * (1 + sin^2(x) + 2cos^2(x))
at x = pi/2, numerator = -1 * (1 + 1^2 + (2)(0)^2) = -2.
denominator is always positive. So f'(x) < 0 in the interval (0, pi/2]
That means f(x) is decreasing in the interval 90, pi/2]

Answer 57

But if you feel like evaluating the whole f'(pi/2), it is no big deal.
The denominator is (1 + sin^2(x))^2 and at x = pi/2 it is +4

Answer 58

f(x) is decreasing in the interval (0, pi/2]

Answer 59

ok I see; thank you.
let me try the other one, then. (if you would please stay, as you can see I do not excel at this...)
ok so I shouldn't worry about the denominator
so what would be a good number to test for [pi, 2pi) ?

Answer 60

midpoint 3/2pi. From the unit circle:
sin(3pi/2) = -1
cos(3pi/2) = 0

You can evaluate the whole f'(x) if you wish including the denominator for the sake of completeness.

Answer 61

so it would be 1(1+1+2(0))?

Answer 62

f'(x) = (-sin(x) * (1 + sin^2(x) + 2cos^2(x) / (1 + sin^2(x))^2

Answer 63

yes, your numerator is correct. You can include the denominator also if you wish.

Answer 64

so it is increasing for [pi, 2pi)!

Answer 65

Yes. That means x = pi is a minimum. You can put x = pi in the original function (not the derivative) and find the y value. That will be your local minimum value of the function.

Answer 66

you said sinpi/2=1 and cospi/2=0 right?
so y= 0/1+1? the minimum is (pi/2,0)?

oh ok x=pi that is what I had been solving on my line
so then y= -1 and (pi, -1) is our minimum?

Answer 67

x = pi is the critical point where the function attains minimum.
You need to put x = pi in the original function.

Answer 68

right so would y= -1 and the minimum be (pi, -1)?

Answer 69

Yes.

Answer 70

so f is decreasing, then increasing and it has a minimum at (pi, -1)
this is my answer?
sorry it took so long for me to get there >_<

Answer 71

I think you are gone, but thank you so very much for your help!
question closing...

Answer 72

To summarize:
The function has a critical point at x = pi in the open interval (0, 2pi)
In the interval (0, pi) the function is decreasing.
At x = pi the function is flat (neither increasing nor decreasing cuz f'(pi) = 0)
In the interval (pi, 2pi) the function is increasing.

Therefore the function has a relative minimum at x = pi and the minimum value is -1.

Answer 73

You are welcome.