Rolle's Theorem has these three hypotheses: f is continuous on [1, 3], f is differentiable on (1, 3) and f(1)=f(3).

The first and the second are obvious for a polynomial function, the third can be easily verified: f(1)=5-12+3=-4, f(3)=5-36+27=-4.

Then there exists at least one c from (1,3) such that...

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Rolle's Theorem has these three hypotheses: f is continuous on [1, 3], f is differentiable on (1, 3) and f(1)=f(3).

The first and the second are obvious for a polynomial function, the third can be easily verified: f(1)=5-12+3=-4, f(3)=5-36+27=-4.

Then there exists at least one c from (1,3) such that f'(c)=0.

f'(x)=-12+6x, this is =0 for x=2 only.

The answer: hypotheses are satisfied, **c=2**.