I know to find the critical numbers you'd find the first derivative and solve for where it is 0, but I just can't get the right answer for this one (Crying)

f(x)= x^(1/5) - x^(-4/5)

Printable View

- Nov 11th 2011, 01:16 PMdeannnaaFind the critical numbers of a function
I know to find the critical numbers you'd find the first derivative and solve for where it is 0, but I just can't get the right answer for this one (Crying)

f(x)= x^(1/5) - x^(-4/5) - Nov 11th 2011, 01:20 PManonimnystefyRe: Find the critical numbers of a function
what did you get for the derivative? maybe the problem is in what you got. :)

- Nov 11th 2011, 01:30 PMdeannnaaRe: Find the critical numbers of a function
f'(x) = 1/[5x^(4/5)] + 4/[5x^(9/5)]

- Nov 11th 2011, 02:34 PManonimnystefyRe: Find the critical numbers of a function
what did you do when you set that equal to zero?

- Nov 11th 2011, 03:44 PMdeannnaaRe: Find the critical numbers of a function
First I factored out 1/5x^4/5

So I had: 1/[5x^(4/5)] x (1 + 4/x)

Then I set that to 0, but the first part is never zero so I just solved for where 1 + 4/x is 0 and got -4. Shouldn't that be the right answer?(Speechless) - Nov 11th 2011, 03:56 PMQuackyRe: Find the critical numbers of a function
$\displaystyle f'(x)=\frac{1}{5x^{\frac{4}{5}}}+\frac{4}{5x^\frac {9}{5}}$

$\displaystyle 0=\frac{1}{5x^{\frac{4}{5}}}+\frac{4}{5x^\frac{9}{ 5}}$

$\displaystyle 0=\frac{1}{5x^{\frac{4}{5}}}(1+\frac{4}{x})$

What is the given answer? Because as far as I can tell, having checked, and having checked again, there's no problem with your work, unless I'm missing something.