Please help with this question

**PapaJones** · May 29th 2012, 11:12 PM

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

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

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

**Siron** · May 29th 2012, 11:22 PM

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

**Goku** · May 29th 2012, 11:41 PM

Derivative of a constant is 0.

**PapaJones** · May 30th 2012, 12:06 AM

anyhelp please?

**Siron** · May 30th 2012, 12:47 AM

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

**Goku** · May 30th 2012, 12:55 AM

It's impossible to solve Or Was the question:

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

**PapaJones** · May 30th 2012, 12:56 AM

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

**Goku** · May 30th 2012, 02:31 AM

Originally Posted by PapaJones

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

Forgot the 6

The derivative of 6 is 0.

**skeeter** · May 30th 2012, 04:37 AM

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.

**draganicimw** · May 30th 2012, 04:46 AM

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

**HallsofIvy** · May 30th 2012, 11:08 AM

Originally Posted by PapaJones

anyhelp please?

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

Thread: Please help with this question