# `f(x) = 2x^2 - 3x + 1, [0,2]` Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers `c` that satisfy the conclusion of the Mean Value Theorem.

Given: `f(x)=2x^2-3x+1,[0,2]`

The function is a continuous polynomial on the closed interval [0,2].

The function is differentiable on the open interval (0,2).

`f'(x)=4x-3`

`f(c)=4c-3`

`[f(2)-f(0)]/(2-0)=[3-1]/2=1`

`4c-3=1`

`4c=4`

`c=1`

The answer is c=1 which is in the interval [0,2].

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