1. Oct 17, 2007

### BuBbLeS01

1. The problem statement, all variables and given/known data
Use Rolle's Theorem to find all values of c n the open interval (a,b) such that f'(c)=0
F(x) = (x^2 - 2x + 3) / (x + 2)
Closed interval [-1, 3]

2. The attempt at a solution
Okay so...
F(-1) = F(3)
F'(x) = [(x+2)(2x+2)] - [(1)(x^2-2x-3)] / (x+2)^2
simplified to (x^2 + 4x - 1) / (x + 2)^2
I need to solve for x and I am having an algebra block! How do I do that?

2. Oct 17, 2007

### Dick

F'(x) vanishes if the numerator vanishes. It's a quadratic equation. But check your derivative first. It didn't come out right.

3. Oct 17, 2007

### BuBbLeS01

Oh okay....One more....
I have (2/3)x^(-1/3) = 1
How do solve for x again?

4. Oct 17, 2007

### Dick

Multiply both sides by 3/2, then take both sides to the power of -3. Where did I come up with 3/2 and -3???

5. Oct 17, 2007

### BuBbLeS01

I don't know??

6. Oct 17, 2007

### Dick

I'm trying to get x by itself. Multiplying by 3/2 cancels the 2/3, taking it to the power of -3 cancels the -1/3 exponent. Now I have x on one side and just numbers on the other. It's called 'algebra'.