SUPPOSE that f [0,2]-> R is continuous on[02] and differentiableon [0,2] and that f(0)=0 f(1)=1, f(2)=1.... show that there exits c1 belongs to (0,1) such that f (c1)=1
Answer:-
Since f : [0,2] -> R is continuous in [0, 2] and differentiable in (0, 2)
So it satisfies Mean Value Theorem.
i.e., ∀ c ∈ ( a, b)
f( 0) = 0
f(1) = 1
From Mean Value Theorem,
∀ c1 ∈ ( 0, 1)
∴ f'(c1) = 1 ∀ c1 ∈ ( 0, 1)
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