Solve Rolle's Theorem: Find c in [-1,3] for F'(c)=0

[BuBbLeS01]

Homework Statement

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?

 

[Dick]

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

 

[BuBbLeS01]

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

 

[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?

 

[BuBbLeS01]

I don't know??

 

[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'.