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# mean value theorem Watch

1. i no how to do part b but part a i am confused any help?
2. Hint: cos(x) from 0 to pi is always decreasing.

Also, do you know the mean value theorem?
3. (Original post by aznkid66)
Hint: cos(x) from 0 to pi is always decreasing.

Also, do you know the mean value theorem?
yh i do in simple terms: f'(c)=(f(b)-f(a))/b-a
4. Okay, so let f(x)=sinx and let c be a point such that a<c<b.

Apply the mean value theorem.

Knowing that the function cos(x) is always decreasing, what can you say about cos(a), cos(c), and cos(b)?
5. so would we have something like this to be working with:

sinb-sina/b-a goes to cosb-cosa/b-a? and then work from there
6. Um...not exactly.

The mean value theorem states that in the bounds from a to b there exists a c such that a<c<b and f'(c)=[f(b)-f(a)]/[b-a].

So let f(x)=sin(x).

According to the mean value theorem, between the a and b values given in the problem, there exists a c such that:

cos(c)=[sin(b)-sin(a)]/[b-a]

Now, what do you know about cos(a), cos(b), and cos(c) given that the function is always decreasing from a to b and a<c<b?
7. cosc is going to lie between cosa and cosb?
8. Yup, but more specifically you know the direction of the inequality.

So now you have:

cos(b)<cos(c)<cos(a)

And:

cos(c)=[sin(b)-sin(a)]/[b-a]

So clearly:

cos(b)<[sin(b)-sin(a)]/[b-a]<cos(a) QED

Do you see what sub in for a and b to do the next part?
9. (Original post by aznkid66)
Yup, but more specifically you know the direction of the inequality.

So now you have:

cos(b)<cos(c)<cos(a)

And:

cos(c)=[sin(b)-sin(a)]/[b-a]

So clearly:

cos(b)<[sin(b)-sin(a)]/[b-a]<cos(a) QED

Do you see what sub in for a and b to do the next part?
i was confused about what the question was asking, so all it wanted was for us to just state that it comes back to what it first stated but by using the MVT?

um i was thinking pi/4 and pi/2 or isit asking us to sub in a function?
10. thank you for the help by the way
11. oh i got it, make b=2x and a=x
12. Well, yes, we were supposed to arrive "back to what it first stated" because we were asked to prove it to be true, not assume it to be true.

Also, I was thinking of different values for a and b. Are you sure the ones you suggested work? ._.

It's quite obviously that a substitution where cosb=cosx, b=...? and cosa=1, a=...? works as a specific case where the statement we just proved implies what we want to deduce.
13. (Original post by aznkid66)
Well, yes, we were supposed to arrive "back to what it first stated" because we were asked to prove it to be true, not assume it to be true.

Also, I was thinking of different values for a and b. Are you sure the ones you suggested work? ._.

It's quite obviously that a substitution where cosb=cosx, b=...? and cosa=1, a=...? works as a specific case where the statement we just proved implies what we want to deduce.
ah sorry jumped the gun, yes b=x and a=1 would work

thank you
14. my bad a=0 i mean

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