# Thread: Rolle's Theorem -- esp. finding derivative of fractions.

1. ## Rolle's Theorem -- esp. finding derivative of fractions.

The ordering and transportation cost C of components used in a manufacturing process is approximated by

$C(x) = 10 [1/x + x/(x+3)]$

where $C$ is measured in thousands of dollars and x is the order of the size in hundreds.

(a) Verify that $C(3) = C(6)$. --done this--
(b) According to Rolle's Theorum, the rate of change of cost must be 0 for some order size in the interval (3,6). Find that order size.

I started b, and trying to find the derivative of the original equation got very hairy very quickly. Any help? Thanks so much in advance.

2. Originally Posted by AppleTini
The ordering and transportation cost C of components used in a manufacturing process is approximated by

$C(x) = 10 [1/x + x/(x+3)]$

where $C$ is measured in thousands of dollars and x is the order of the size in hundreds.

(a) Verify that $C(3) = C(6)$. --done this--
(b) According to Rolle's Theorum, the rate of change of cost must be 0 for some order size in the interval (3,6). Find that order size.

I started b, and trying to find the derivative of the original equation got very hairy very quickly. Any help? Thanks so much in advance.
How about posting what you did so we can look at it? You have
C= 10[1/x+ x/(x+3)]. You can most easily differentiate 1/x by writing it as $x^{-1}$. x/(Ix+3) requires the quotient rule.

3. Does the whole thing need the product rule? If so..

10[ $-x^{-2}$ + 2x+3 / $(x+3)^{2}$]

4. Originally Posted by AppleTini
The ordering and transportation cost C of components used in a manufacturing process is approximated by

$C(x) = 10 [1/x + x/(x+3)]$

where $C$ is measured in thousands of dollars and x is the order of the size in hundreds.

(a) Verify that $C(3) = C(6)$. --done this--
(b) According to Rolle's Theorum, the rate of change of cost must be 0 for some order size in the interval (3,6). Find that order size.

I started b, and trying to find the derivative of the original equation got very hairy very quickly. Any help? Thanks so much in advance.
Do you mean $C(x) = 10 \left( \frac{1}{x} + \frac{x}{x+3}\right)$ ?

5. Originally Posted by mr fantastic
Do you mean $C(x) = 10 \left( \frac{1}{x} + \frac{x}{x+3}\right)$ ?
Yes, this. I struggle with the "Math" button.

6. Originally Posted by AppleTini
Yes, this. I struggle with the "Math" button.
$
{\text{How }}C\left( 3 \right) = C\left( 6 \right) \Rightarrow \exists \xi \in \left( {3,6} \right){\text{ in such case }}C'\left( \xi \right) = 0
$

And this respond (b) because $
C'\left( \xi \right)
$
is the rate of change

7. $C(x) = 10\left(\frac{1}{x} + \frac{x}{x+3}\right)$

$C(x) = 10\left(\frac{x^2+x+3}{x^2+3x}\right)$

$C'(x) = 10\left(\frac{(x^2+3x)(2x+1) - (x^2+x+3)(2x+3)}{(x^2+3x)^2}\right)$

for $C'(x) = 0$ ...

$(x^2+3x)(2x+1) = (x^2+x+3)(2x+3)$

$2x^3 + 7x^2 + 3x = 2x^3 + 5x^2 + 9x + 9$

$2x^2 - 6x - 9 = 0$

$(2x - 9)(x + 1) = 0$

$x = \frac{9}{2}$ is in the interval

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# the ordering and transportation cost c of the components used in manufacturing a product is

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