# Please help with this question

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• May 29th 2012, 11:12 PM
PapaJones
Please 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 PM
Siron
Re: Please help with this question
The derivative $L'(x)$ is not correct.
• May 29th 2012, 11:41 PM
Goku
Re: Please help with this question
Derivative of a constant is 0.
• May 30th 2012, 12:06 AM
PapaJones
Re: Please help with this question
anyhelp please?
• May 30th 2012, 12:47 AM
Siron
Re: Please help with this question
You have to solve $L'(x)=0 \Leftrightarrow \frac{-64000}{x^3}=0$, this is impossible
• May 30th 2012, 12:55 AM
Goku
Re: Please help with this question
It's impossible to solve Or Was the question:

$L(x) = \frac{32000}{x^2+6}$
• May 30th 2012, 12:56 AM
PapaJones
Re: Please help with this question
Quote:

Originally Posted by Siron
You have to solve $L'(x)=0 \Leftrightarrow \frac{-64000}{x^3}=0$, this is impossible

L'(x) = 0 = -64000/x^3 +6

Forgot the 6
• May 30th 2012, 02:31 AM
Goku
Re: Please help with this question
Quote:

Originally Posted by PapaJones
L'(x) = 0 = -64000/x^3 +6

Forgot the 6

The derivative of 6 is 0.
• May 30th 2012, 04:37 AM
skeeter
Re: Please help with this question
Quote:

Originally Posted by PapaJones
if L(x)=32000/x^2+6

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

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

is the original function $L(x) = \frac{32000}{x^2} + 6$ or $L(x) = \frac{32000}{x^2+6}$ ?

... in either case, your derivative is incorrect.
• May 30th 2012, 04:46 AM
draganicimw
Re: 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 AM
HallsofIvy
Re: Please help with this question
Quote:

Originally Posted by PapaJones
anyhelp please?

What kind of "help" do you want? You have simply ignored every post so far.