Mean Value Theorem Q

[Luckeh!!!!]

pic

I get this

obviously I'm missing something, but what?

[Anu]

(sin(c))^2*cos(c) < 1 => (sin(b))^3 < 3b

[Luckeh!!!!]

how do you work out that it is f'(c) is < 1 ?

[Anu]

You've got:
(sin(b))^3 = 3b*(sin(c))^2*cos(c)

Now, look at the sine-cosine circle. And it will tell you that,
(sin(c))^2*cos(c) < 1

Hence,
(sin(b))^3 < 3b

[Luckeh!!!!]

Originally Posted by Anu

You've got:
(sin(b))^3 = 3b*(sin(c))^2*cos(c)

Now, look at the sine-cosine circle. And it will tell you that,
(sin(c))^2*cos(c) < 1

Hence,
(sin(b))^3 < 3b

I dont have 3b*(sin(c))^2*cos(c)

i have 3b*(sin(b))^2*cos(b) though :/

that doesn't make a difference tohugh does it? or have I done another thing wrong?

[Anu]

Originally Posted by Luckeh!!!!

I dont have 3b*(sin(c))^2*cos(c)

i have 3b*(sin(b))^2*cos(b) though :/

that doesn't make a difference tohugh does it? or have I done another thing wrong?

Look in the middle of your notes, where it says

3(sin(c))^2*cos(c) = ((sin(b))^3 - 0)/(b-0)

This gives

3b*(sin(c))^2*cos(c) = (sin(b))^3

If you're not familiar with the unit circle (which you should be), here's a link from Mathworld

Unit Circle