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

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

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

The derivative $L'(x)$ is not correct.

Derivative of a constant is 0.

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

It's impossible to solve Or Was the question:

$L(x) = \frac{32000}{x^2+6}$

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

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

Forgot the 6
The derivative of 6 is 0.

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.

If L(x) =32000/(x^2+6) the derivation is:
L'(x) = - 64000/[(x^2+6)^2]*2x and L'(x) = 0 <=> x = 0

Originally Posted by PapaJones