if L(x)=32000/x^2+6

and L'(x)=-64000/x^3+6

Solve L'(x)=0 (to two decimal places)

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- May 29th 2012, 11:12 PMPapaJonesPlease help with this question
if L(x)=32000/x^2+6

and L'(x)=-64000/x^3+6

Solve L'(x)=0 (to two decimal places) - May 29th 2012, 11:22 PMSironRe: Please help with this question
The derivative $\displaystyle L'(x)$ is not correct.

- May 29th 2012, 11:41 PMGokuRe: Please help with this question
Derivative of a constant is 0.

- May 30th 2012, 12:06 AMPapaJonesRe: Please help with this question
anyhelp please?

- May 30th 2012, 12:47 AMSironRe: Please help with this question
You have to solve $\displaystyle L'(x)=0 \Leftrightarrow \frac{-64000}{x^3}=0$, this is impossible

- May 30th 2012, 12:55 AMGokuRe: Please help with this question
It's impossible to solve Or Was the question:

$\displaystyle L(x) = \frac{32000}{x^2+6}$ - May 30th 2012, 12:56 AMPapaJonesRe: Please help with this question
- May 30th 2012, 02:31 AMGokuRe: Please help with this question
- May 30th 2012, 04:37 AMskeeterRe: Please help with this question
- May 30th 2012, 04:46 AMdraganicimwRe: Please help with this question
If L(x) =32000/(x^2+6) the derivation is:

L'(x) = - 64000/[(x^2+6)^2]*2x and L'(x) = 0 <=> x = 0 - May 30th 2012, 11:08 AMHallsofIvyRe: Please help with this question