The function f (x) = x3-x2 + x2 + 14 is known. It is proved that there exists x0 ∈ (0,12) such that f (x0) = x0

It is proved that: Let G (x) = f (x) - X. ∵ g (0) = 14, G (12) = f (12) - 12 = - 18, ∵ g (0) · g (12) < 0. And the function g (x) is continuous on [0,12], so there exists x0 ∈ (0,12), so that G (x0) = 0. That is, f (x0) = x0

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