prove for k is bigger than or equal to 0. on attachment!
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Originally Posted by ardam prove for k is bigger than or equal to 0. on attachment! Consider $\displaystyle f(x) = x^{1/3}$. Then $\displaystyle f(x + h) \leq f(x) + h f'(x)$ for $\displaystyle h > 0$ and $\displaystyle x > 0$.
Last edited by mr fantastic; Jan 13th 2009 at 11:20 AM. Reason: Added x > 0
Originally Posted by mr fantastic Consider $\displaystyle f(x) = x^{1/3}$. Then $\displaystyle f(x + h) \leq f(x) + h f'(x)$ for $\displaystyle h > 0$. Smart alec! LOL. I knew I was missing something obvious. I tried so many algebraic manipulations it was driving me nuts! Hahaha
Last edited by mr fantastic; Jan 13th 2009 at 11:20 AM. Reason: I had a posterior reason
Originally Posted by mr fantastic Consider $\displaystyle f(x) = x^{1/3}$. Then $\displaystyle f(x + h) \leq f(x) + h f'(x)$ for $\displaystyle h > 0$. thanks for help but not sure why you differentiate.
Last edited by mr fantastic; Jan 13th 2009 at 11:20 AM.
Originally Posted by ardam thanks for help but not sure why you differentiate. It's a standard inequality. Use $\displaystyle f(x) = x^{\frac{1}{3}}$ and $\displaystyle h = 1$. When you sub everything into the inequality you should get $\displaystyle (x + 1)^{\frac{1}{3}} \leq x^{\frac{1}{3}} + \frac{1}{3}x^{-\frac{2}{3}}$.
Last edited by mr fantastic; Jan 13th 2009 at 11:21 AM.
Originally Posted by Prove It It's a standard inequality. Use $\displaystyle f(x) = x^{\frac{1}{3}}$ and $\displaystyle h = 1$. When you sub everything into the inequality you should get $\displaystyle (x + 1)^{\frac{1}{3}} \leq x^{\frac{1}{3}} + \frac{1}{3}x^{-\frac{2}{3}}$. oh right yer i previously got that far but not sure wot technique to use to prove.
Originally Posted by ardam oh right yer i previously got that far but not sure wot technique to use to prove. Substitute $\displaystyle x = k$ and make the obvious re-arrangement.
Originally Posted by mr fantastic Substitute $\displaystyle x = k$ and make the obvious re-arrangement. so then you prove its bigger than of equal to 0
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 $\displaystyle \sqrt[3]{k}$ to the other side. What do you get?
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 $\displaystyle \sqrt[3]{k}$ to the other side. What do you get? you get that all the terms are less than 0
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.
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