Question

Help Please!!(:
Given two real numbers a and b, with a 12 years ago

Answer 1

Do you know the mean value theorem?

Answer 2

yes. \[f'(c)= \frac{ f(b)-f(a) }{ b-a }\]

Answer 3

do you know it is after 1 am?

Answer 4

Nice. Let f(x)=sin(x). What do you get then?

Answer 5

I still got 50 mins till 1 am :P

Answer 6

I'm from Hawaii lol

Answer 7

night night

Answer 8

sin(b)-sin(a) / b-a

Answer 9

Let f(x)=sin(x), and throw some absolute value signs on for good measure:\[\left|\cos (c)\right|=\frac{\left|\sin(b)-\sin(a)\right|}{\left|b-a\right|}\]The cosine is there because that is the derivative of sine. Everything make sense so far?

Answer 10

yes!

Answer 11

Sweet. Now keep your eyes on the prize. We want to show that:\[\left|\sin(b)-\sin(a)\right|\le \left|b-a\right|\]Since a < b, we can divide both sides of this by |b - a| to obtain:\[\frac{\left|\sin(b)-\sin(a)\right|}{\left|b-a\right|}\le 1\]Do any ideas come to mind?

Answer 12

I kinda got lost. Where did the 1 come from?

Answer 13

oh wait nvm haha

Answer 14

Ah sry, I'll show my work. First we start with:\[\left|\sin(b)-\sin(a)\right|\le \left|b-a\right|\]Now I'm going to divide both sides by |b-a|:\[\frac{\left|\sin(b)-\sin(a)\right|}{|b-a|}\le \frac{\left|b-a\right|}{|b-a|}\] The right hand side becomes a 1. Thats where it came from.

Answer 15

I caught my mistake, but thanks.. so is that where it ends? or is there more to the problem?

Answer 16

I know that when using the MVT the value is less then or equal to 1

Answer 17

We havent shown anything yet. All we have is that:\[\frac{|\sin(b)-\sin(a)|}{|b-a|}=|\cos(c)|\]But we want:\[\frac{|\sin(b)-\sin(a)|}{|b-a|}\le 1\]There's one more thing that needs to be said about |cos(c)|.

Answer 18

it's greater or equal tot he function?

Answer 19

What is the biggest value that |cos(c)| can take?

Answer 20

1?

Answer 21

or anything greater

Answer 22

Thats right, 1, is the biggest value. So we have this:\[\frac{|\sin(b)-\sin(a)|}{|b-a|}=|\cos(c)|\]and \[|\cos(c)|\le 1\]Putting these two together yields:\[\frac{|\sin(b)-\sin(a)}{|b-a|}\le 1\]and if we multiply both sides by |b-a|, we get:\[|b-a|\cdot \frac{|\sin(b)-\sin(a)|}{|b-a|}\le 1\cdot |b-a|\]\[\Longrightarrow |\sin(b)-\sin(a)|\le |b-a|\]

Answer 23

Wow, That wasn't as complicated. I guess I over thinked the whole process. haha thank you soooo much! I really approcoate all the help! (: @joemath314159