Simple Prime Question

**Brazuca** · July 6th 2009, 06:49 PM

F(X) = (2/3)(X-2)^(-1/3)

F'(X)= (-2/9)(X-2)^(-4/3)

F''(X) = (8/27)(X-2)^(-5/3) Is F''(X) correct???

**Rapha** · July 6th 2009, 06:50 PM

Hi.

Originally Posted by Brazuca

F(X) = (2/3)(X-2)^(-1/3)

F'(X)= (-2/9)(X-2)^(-4/3)

F''(X) = (8/27)(X-2)^(-5/3) Is F''(X) correct???

No!

I think it is a typo, but it is

$F''(X) = (8/27)(X-2)^{\displaystyle-(4/3) - 1}= (8/27)(X-2)^{\displaystyle-4/3 - 3/3} = (8/27)(X-2)^{\displaystyle-7/3}$

(F'(X) is correct!)

Yours
Rapha

**Brazuca** · July 6th 2009, 06:55 PM

Wasn't a typo. I just had it incorrect.

Thanks now I understand how it works.

**Brazuca** · July 6th 2009, 07:24 PM

I have two more questions involving this problem.

F(X) = (2/3)(X-2)^(-1/3) when X = 1 why is the final answer negative?

F'(X)= (-2/9)(X-2)^(-4/3) when X = 1 why is the final answer negative?

**Rapha** · July 6th 2009, 07:41 PM

Originally Posted by Brazuca

I have two more questions involving this problem.

F(X) = (2/3)(X-2)^(-1/3) when X = 1 why is the final answer negative?

Just substitute X=1, then

$F(1) = \frac{2}{3}*(1-2)^{-1/3} = \frac{2}{3}*(-1)^{-1/3}$

We know that $a^{-1} = \frac{1}{a}$

Therefor

$F(1) = \frac{2}{3}*(-1)^{-1/3} = \frac{2}{3}*\frac{1}{\displaystyle(-1)^{1/3}}$

Furthermore we know that $x^{1/2} = \sqrt{x}$ and $x^{1/3} = \sqrt[3]{x}$

Thus

$F(1) = \frac{2}{3}*\frac{1}{\displaystyle(-1)^{\displaystyle1/3}} = \frac{2}{3}*\frac{1}{\sqrt[3]{-1}} = \frac{2}{3}*(-1) = - 2/3 < 0$

Originally Posted by Brazuca

F'(X)= (-2/9)(X-2)^(-4/3) when X = 1 why is the final answer negative?

Same here

$F'(1) = (-2/9) (1-2)^{-4/3} = -2/9*(-1)^{\displaystyle-4/3} = -2/9 * \frac{1}{(-1)^{\displaystyle4/3}}$

$= -2/9 * \frac{1}{\displaystyle((-1)^4)^{\displaystyle1/3}} = -2/9*\frac{1}{1^{\displaystyle1/3}} = -2/9 * \frac{1}{\sqrt[3]{1}} = -2/9*1 = -2/9 <0$

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