Question

Show that f satisfies the hypotheses of Rolle's theorem on [a,b] and find all numbers c in (a,b) such that f'(c) = 0.

f(x) = sin 2x; [0, pi]

Answer 1

f(0) = f(pi) = 0;
f'(x) = cos 2x * 2 = 2 cos 2x;

Answer 2

cos x = 0 at pi/2 and 3pi/2
f(pi) - f(0) / pi - 0 = 0 = 2 cos 2x

Answer 3

yeh thats easy

Answer 4

Sadly, I got pi/2 and 3pi/2 when it should be pi/4 and 3pi/4

Answer 5

Is that because I didn't account for the 2 before the cos 2x?

Answer 6

since you divide it by 2?

Answer 7

yes

Answer 8

Sorry, you've helped me out before. I may be hopeless, but I'm trying HARD as of late.

Answer 9

That's all there is to it? Divide it by 2 to account for the 2 before the cos 2x? Wow.

Answer 10

Good answer and thanks for the help!

Answer 11

yes you have cos(u)=0 on interval [0,pi] when u=pi/2
cos(pi/2)=0

so we want to find when cos(2x)=cos(pi/2)

so since the outside functions are the same
the inside functions have to be the same
so we have 2x=pi/2
solve for x

Answer 12

Ah wait! So I don't account for the 2 BEFORE the cos 2x, but rather the 2 BEFORE the x?

Answer 13

rolles theorm , basically common sense, like intermediate value theorm, why does he deserve a theorm named after him

Answer 14

lol
mean value thm is better

Answer 15

myininaya, was my previous question correct?

Answer 16

right you have 2cos2x=0
divide both sides by 2
you have cos2x=0

Answer 17

Yeah, it definitely is. Using common sense! :) Thanks guys (and gals)!

Answer 18

0=cos(pi/2)
so cos(2x)=cos(pi/2)

Answer 19

thats because its the general case of rolle, but still when I first heard learnt bout it last year in uni it wasnt hard, seems like common sense.

Answer 20

Yeah, it's making a bunch more sense. The 2 BEFORE cos doesn't affect it at all if it's equal to 0!

Answer 21

right

Answer 22

elecengineer, it may be making a bunch more sense to you than the common man. :P Even though I'm a programmer, math isn't really in my genes and the theories we've covered the past 2 years -- while dabbling in high-end mathematics -- don't seem quite as hard covered in a programming class compared to the stuff we're doing in Calculus with no motivating factor. :P

Answer 23

Thanks myininaya, it's making a "lotta" sense!

Answer 24

i dunno how to dis but u should b a fan of me :)

Answer 25

Never a fan of someone typing broken English and whoring for some sort of advancement. Thanks myininaya (and elecengineer)!

Answer 26

yeh well I hate programming :)

I had to two computing subjects last, nearly killed me lol, probably not but yeh I dont have the patience for that.

Answer 27

sometimes my english is broken lol

Answer 28

If you have to take another, let me know. That's one thing I can definitely help you with. That, plus losing weight / gaining muscle / becoming an athlete. :P Prior to becoming a computer geek, my goal was to train athletes by opening up a gym to compete against guys like Joe DeFranco.

Answer 29

but we all have our pet peeves

Answer 30

yeh but its the net, and yeh im not all that good at english but I figure people can understand it , so theres not much point making everything look good.

Answer 31

Nah, it just reminds me of people on computer forums (yeah, I'm a serious geek!) who dole out tiny bits of advice and want reputation points, which is what I assume "rosey" wanted. Speaking "quick English" isn't a pet peeve, speaking BROKEN English is.

Answer 32

Both of you have perfectly understandable English. rosey didn't, but I assume she CAN write proper English, she was just too busy begging to be "followed" or given a Good Answer or whatever it was she was seeking.

Anyways, thanks for the help! I'm out for the night (big exam!) and hopefully won't have to come begging ya'll for help anymore. Hehe.

Answer 33

Um I wasn't speaking broken english babe but if that's how you feel about me I don't really care to have you as a fan anyways.

Answer 34

<3

Answer 35

i dunno how to dis but u should b a fan of me :)

Like I said, I'm out. You can try to decipher what it was you're trying to say and re-phrase it tomorrow, "babe". Peace out guys!

Answer 36

good luck

Answer 37

You know that the MVT is just Rolle's theorem transformed.

Answer 38

but rolles thm only works for f(a)=f(b)

while mean value them works for that and more

Answer 39

To prove the mean value theorem you can just setup a transformation so that it is the form where you can apply the rolle's theorem.

Answer 40

assuming f is continuous and differentiable of course

Answer 41

hey alchemista will you show me on a different thread
i will make a new one