finding minimum

**el123** · September 1st 2009, 06:17 PM

$AC=\frac{TC}{x}=\frac{2000}{x}+20+\frac{10}{\sqrt{ x}}$
So i have to show that there is no minimum.

thus i have to differentiate the function, which apparently gives this

$\frac{dAC}{dx}=AC'= -\frac{2000}{x^2}-\frac{5}{\sqrt[3]{}x^3}$

I don't know how they got that answer , can someone please explain?

**skeeter** · September 1st 2009, 06:31 PM

Originally Posted by el123

$AC=\frac{TC}{x}=\frac{2000}{x}+20+\frac{10}{\sqrt{ x}}$
So i have to show that there is no minimum.

thus i have to differentiate the function, which apparently gives this

$\frac{dAC}{dx}=AC'= -\frac{2000}{x^2}-\frac{5}{\sqrt[3]{}x^3}$

I don't know how they got that answer , can someone please explain?

$AC = \frac{2000}{x}+20+\frac{10}{\sqrt{x}}$

$AC = 2000x^{-1} + 20 + 10x^{-\frac{1}{2}}$

use the power rule ...

$AC' = -2000x^{-2} - 5x^{-\frac{3}{2}}$

rewrite ...

$AC' = -\frac{2000}{x^2}-\frac{5}{\textcolor{red}{\sqrt{x^3}}}$

note there is no "cube" root in the denominator of the last term.

**el123** · September 1st 2009, 06:37 PM

Thanks.

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