The function f (x) is differentiable on [0,1] and 0

Let f (x) = f (x) - X,
If f (x) = x has a unique real root in (0,1), then f (x) = 0 has a solution
F(X)′=f′(x)-1
When ∵ f ′ (x) ≠ 1, ∵ f ′ (x) > 0, or F ′ (x) 0,
F(0)=f(0)-0=f(0)>0
F(1)=f(1)-1

RELATED INFORMATIONS

1. Let f (x) be differentiable on [0,1] and 0
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