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

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Homework Help Overview

The discussion revolves around applying Rolle's Theorem to find values of c in the interval [-1, 3] such that f'(c) = 0 for the function F(x) = (x^2 - 2x + 3) / (x + 2).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to verify the conditions of Rolle's Theorem by checking if F(-1) equals F(3) and calculating the derivative F'(x). Some participants question the correctness of the derivative calculation and suggest reviewing it. There is also a mention of solving a different equation involving exponents, leading to confusion about the steps taken.

Discussion Status

The discussion is ongoing, with participants exploring the derivative and its implications. Some guidance has been offered regarding algebraic manipulation, but there is no explicit consensus on the correctness of the derivative or the approach to solving for x.

Contextual Notes

Participants express uncertainty about their algebraic manipulations and the steps involved in solving the equations presented. There is a focus on ensuring the derivative is calculated correctly before proceeding.

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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?
 
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F'(x) vanishes if the numerator vanishes. It's a quadratic equation. But check your derivative first. It didn't come out right.
 
Oh okay...One more...
I have (2/3)x^(-1/3) = 1
How do solve for x again?
 
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?
 
I don't know??
 
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'.
 

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