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

$\displaystyle L(x) = \frac{32000}{x^2+6}$
• May 30th 2012, 12:56 AM
PapaJones
Quote:

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

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

Forgot the 6
• May 30th 2012, 02:31 AM
Goku
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
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 $\displaystyle L(x) = \frac{32000}{x^2} + 6$ or $\displaystyle L(x) = \frac{32000}{x^2+6}$ ?

... in either case, your derivative is incorrect.
• May 30th 2012, 04:46 AM
draganicimw