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 30th 2012, 12:12 AMPapaJonesPlease 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 30th 2012, 12:22 AMSironRe: Please help with this question
The derivative is not correct.

- May 30th 2012, 12:41 AMGokuRe: Please help with this question
Derivative of a constant is 0.

- May 30th 2012, 01:06 AMPapaJonesRe: Please help with this question
anyhelp please?

- May 30th 2012, 01:47 AMSironRe: Please help with this question
You have to solve , this is impossible

- May 30th 2012, 01:55 AMGokuRe: Please help with this question
It's impossible to solve Or Was the question:

- May 30th 2012, 01:56 AMPapaJonesRe: Please help with this question
- May 30th 2012, 03:31 AMGokuRe: Please help with this question
- May 30th 2012, 05:37 AMskeeterRe: Please help with this question
- May 30th 2012, 05:46 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, 12:08 PMHallsofIvyRe: Please help with this question