# Thread: h/w due for 2morrow

1. ## h/w due for 2morrow

prove for k is bigger than or equal to 0.

on attachment!

2. Originally Posted by ardam
prove for k is bigger than or equal to 0.

on attachment!
Consider $f(x) = x^{1/3}$.

Then $f(x + h) \leq f(x) + h f'(x)$ for $h > 0$ and $x > 0$.

3. Originally Posted by mr fantastic
Consider $f(x) = x^{1/3}$.

Then $f(x + h) \leq f(x) + h f'(x)$ for $h > 0$.
Smart alec! LOL. I knew I was missing something obvious. I tried so many algebraic manipulations it was driving me nuts! Hahaha

4. Originally Posted by mr fantastic
Consider $f(x) = x^{1/3}$.

Then $f(x + h) \leq f(x) + h f'(x)$ for $h > 0$.
thanks for help but not sure why you differentiate.

5. Originally Posted by ardam
thanks for help but not sure why you differentiate.
It's a standard inequality.

Use $f(x) = x^{\frac{1}{3}}$ and $h = 1$.

When you sub everything into the inequality you should get

$(x + 1)^{\frac{1}{3}} \leq x^{\frac{1}{3}} + \frac{1}{3}x^{-\frac{2}{3}}$.

6. Originally Posted by Prove It
It's a standard inequality.

Use $f(x) = x^{\frac{1}{3}}$ and $h = 1$.

When you sub everything into the inequality you should get

$(x + 1)^{\frac{1}{3}} \leq x^{\frac{1}{3}} + \frac{1}{3}x^{-\frac{2}{3}}$.
oh right yer i previously got that far but not sure wot technique to use to prove.

7. Originally Posted by ardam
oh right yer i previously got that far but not sure wot technique to use to prove.
Substitute $x = k$ and make the obvious re-arrangement.

8. Originally Posted by mr fantastic
Substitute $x = k$ and make the obvious re-arrangement.
so then you prove its bigger than of equal to 0

9. Originally Posted by ardam
so then you prove its bigger than of equal to 0
This inequality is known to be true.

So make the appropriate substitutions and rewrite the fractional powers as surds, then move the $\sqrt[3]{k}$ to the other side.

What do you get?

10. Originally Posted by Prove It
This inequality is known to be true.

So make the appropriate substitutions and rewrite the fractional powers as surds, then move the $\sqrt[3]{k}$ to the other side.

What do you get?
you get that all the terms are less than 0

11. Originally Posted by ardam
you get that all the terms are less than 0
Surely you can re-arrange the inequality I gave you (and there's a small edit I made) into the form of the inequality you're required to prove. This is the only bit of the solution that has been left for you to do.