Question

find any extrema for sin(x)-cos(x) on (0,2Pi)
I already got the second derivative which is -sin(x)cos(x), where do I go from here?

Answer 1

what did you get for the first derivative?

Answer 2

I got F(x)=cos(x)+sin(x)

Answer 3

ok did you set that equal to 0 yet to find the critical numbers?

Answer 4

Well, thats where I am running into a problem. It has been a while since I did calc and after I set it to zero, I don't know how to solve it.

Answer 5

so we have cos(x)+sin(x)=0
which means we are looking for when cos(x) and sin(x) have opposite values since cos(x)+sin(x)=0
means cos(x)=-sin(x)

Answer 6

recall at pi/4 cos and sin are exactly the same
can you tell me an example where they are opposite

Answer 7

http://www.google.com/imgres?imgurl=http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/720px-Unit_circle_angles_color.svg.png&imgrefurl=http://en.wikipedia.org/wiki/Unit_circle&h=225&w=225&tbnid=NHFRVspjd-vZrM:&zoom=1&tbnh=186&tbnw=186&usg=__sOrTVeIoKl48_TodEan8dIx7UGk=&docid=D09vYD7H2hKF-M&itg=1&sa=X&ei=pgBCU5KqHeSw2wXa-4HABw&sqi=2&ved=0CJsBEPwdMAo
this could be helpful

Answer 8

5pi/4?

Answer 9

I am just guessing here

Answer 10

at 5pi/4 they are exactly the same

Answer 11

i'm asking you to look at that unit circle and use the definition of opposite

Answer 12

i will give you a hint
look at the 2nd and 4th quadrants

Answer 13

Ahh, I see what you mean. 3pi/4?

Answer 14

that is one
there is one more answer

Answer 15

and 7pi/4?

Answer 16

so the critical numbers are x=3pi/4 , x=7pi/4

Answer 17

between 0 and 2pi anyways

Answer 18

Does it matter any that the problem asks me to use the second derivative?

Answer 19

second derivative test*

Answer 20

also we don't have any extrema values at the endpoints because our endpoints aren't actually included since you have ( ) instead of [ ]

Answer 21

you could use the second derivative to determine if they are relative max or relative min
or neither

Answer 22

f''>0 means it is concave up and you have a relative min
example: f=x^2 is concave up and has a relative min and f''=2>0

f''<0 means it is concave down and you have a relative max
example f=-x^2 is concave down and has a relative max and f''=-2<0

f''=0 means it is neither relative max or min
example f=x^3; f'=3x^2 critical numbers are x=0 f''=6x f''(0)=0 <-- no relative max or min for f=x^3

Answer 23

anyways your second derivative you are missing a plus sign somewhere

Answer 24

and no relative min or max means no extrema, right?

Answer 25

yea, thats a typo

Answer 26

at that particular value yes

Answer 27

so, to answer my problem, I would solve it with the first derivative, then how would I check it with the second derivative?

Answer 28

cause my teacher wants me to impliment the second derivative test

Answer 29

also local (aka relative) extrema do not occur at endpoints

after you find when f'=0 use f'' to determine if you have relative extrema

Answer 30

plug in your critical numbers into your second derivative

Answer 31

and use that paragraph i gave you above to help you determine what kinda extrema you have

Answer 32

So, I would have -sin(pi/4) + cos(pi/4) =0 , then do the same for x=7pi/4?

Answer 33

err why are you pluggin in pi/4?

Answer 34

that was one of my critical numbers, right?

Answer 35

nope

Answer 36

"so the critical numbers are x=3pi/4 , x=7pi/4"

Answer 37

recall we were looking for when cos(x) and sin(x) were oppsites

Answer 38

yes

Answer 39

So, if they are not the critical numbers, what are they?

Answer 40

...

Answer 41

I'm sorry, I am slow today and I am sure it is testing your patience.

Answer 42

i think you need to read above
i already told you the critical numbers and you even quoted me on it so i don't understand

Answer 43

well, I am a little confused, because I thought you were suposed to substitute the critical numbers into the second derivative, right?

Answer 44

yep but pi/4 was never a critical number

Answer 45

so i asked you why you were plugging in that value

Answer 46

pi/4 is not equal to 3pi/4 or 7pi/4

Answer 47

so, is 7pi/4 the only critical number?

Answer 48

why?

Answer 49

we said earlier 3pi/4, 7pi/4 are critical numbers

Answer 50

they were critical numbers because they made f'=0

Answer 51

Alright, I think I got it now. Thank you for your help and patience.

Answer 52

do you understand why I asked you why you were plugging in pi/4?

Answer 53

ya, because it wasn't a critical number, right?

Answer 54

the only numbers you should be plugging in are the critical numbers which were 3pi/4 , 7pi/4

Answer 55

Alright, so plug those in and solve them for zero?

Answer 56

no solving for zero

Answer 57

plug in then use
f''>0 means it is concave up and you have a relative min
example: f=x^2 is concave up and has a relative min and f''=2>0

f''<0 means it is concave down and you have a relative max
example f=-x^2 is concave down and has a relative max and f''=-2<0

f''=0 means it is neither relative max or min
example f=x^3; f'=3x^2 critical numbers are x=0 f''=6x f''(0)=0 <-- no relative max or min for f=x^3