if L(x)=32000/x^2+6
and L'(x)=-64000/x^3+6
Solve L'(x)=0 (to two decimal places)
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if L(x)=32000/x^2+6
and L'(x)=-64000/x^3+6
Solve L'(x)=0 (to two decimal places)
The derivativeis not correct.
Derivative of a constant is 0.
anyhelp please?
You have to solve, this is impossible
It's impossible to solve Or Was the question:
L'(x) = 0 = -64000/x^3 +6Quote:
Originally Posted by Siron
You have to solve
Forgot the 6
The derivative of 6 is 0.Quote:
Originally Posted by PapaJones
is the original functionQuote:
Originally Posted by PapaJones
or
?
... 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
What kind of "help" do you want? You have simply ignored every post so far.Quote:
Originally Posted by PapaJones