# Math Help - Please help with this question

1. ## 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)

2. ## Re: Please help with this question

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

3. ## Re: Please help with this question

Derivative of a constant is 0.

4. ## Re: Please help with this question

anyhelp please?

5. ## Re: Please help with this question

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

6. ## Re: Please help with this question

It's impossible to solve Or Was the question:

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

7. ## Re: Please help with this question

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

8. ## Re: Please help with this question

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

Forgot the 6
The derivative of 6 is 0.

9. ## Re: Please help with this question

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.

10. ## 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

11. ## Re: Please help with this question

Originally Posted by PapaJones
anyhelp please?
What kind of "help" do you want? You have simply ignored every post so far.