How do you find a number c that satisfies the conclusion of the theorem for the function $f(x) = x^2 - 3x + 1$ on the interval [-1,1]?

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Answer 1 · 2015-09-12T10:30:31

$c$ is a solution to the equation $f'(x) = (f(1)-f(-1))/(1-(-1))$

For $f(x) = x^2 - 3x + 1$ on the interval [-1,1]

We get:

$f'(x) = (f(1)-f(-1))/(1-(-1))$

$2x-3 = ((-1)-(5))/(1-(-1)) = (-6)/2 = -3$

So $x=0$

This is the only solution and is, of course in the interval mentioned in the conclusion of MVT. That is, it is in $(-1,1)$ .

So the $c$ we want is $c=0$ .

How do you find a number c that satisfies the conclusion of the theorem for the function f(x) = x^2 - 3x + 1f(x) = x^2 - 3x + 1 on the interval [-1,1]?