Question

Working on Rolle's Theorem right now:
C(x) = 10(1/x + (x/(x+3)))
Find the derivative and solve for x in the derivative.

Answer 1

@RadEn Can you help?

Answer 2

To have a question on Rolle's Theorem you need two points at which the function has the same value, then show that f'(x) has a value of zero at some intermediate point.

Answer 3

Ok, I solved C(3) = C(6)

Answer 4

OK. Now calculate C'(x), and show that it has a zero between 3 and 5.

Answer 5

That's the part I'm having trouble with. I keep messing up somehow because when I solve for it, I get imaginary numbers

Answer 6

\[C'(x) = 10[-1/x^2 - x^2/(x+3)^2]\]
That's the derivative I get

Answer 7

Is the function truly\[\frac{ 10 }{ \left( \frac{ 1 }{ x }+\frac{ x }{ x+3 } \right) }\]?

Answer 8

No, it's the denominator times the numerator

Answer 9

Opps, OK. I'm calculating the derivative.

Answer 10

C(x) = 10(1/x + (x/(x+3))

You failed to apply the quotient rule to the second term in parentheses.The whole derovative should be:

\[10(\frac{ -1 }{ x ^{2} }+\frac{ 3 }{ \left( x+3)^{2} \right) })\] should be

Answer 11

Ok, I'm working it out again now

Answer 12

Alright, I got how you did the derivative, but now I'm having trouble solving for x.

Answer 13

I got to \[1/3 = x^2/(x+3)^2\]

Answer 14

Get a common denominator for the parenthetical term, and set the numerator( it's a quadratic) to zero. The Quadratic Formula will give you two zeros: one of them is between 3 and 6.

Answer 15

Ok, thank you. I'll work it out

Answer 16

Let me know if you can't get ~4.098

Answer 17

Ok

Answer 18

I'm not getting that

Answer 19

Actually, nevermind, I got it :D

Answer 20

Thanks! :)