Answer to Question #246209 in Real Analysis for tannu

Question #246209

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

1
Expert's answer
2021-10-26T16:44:44-0400

Answer:-

Since f : [0,2] -> R is continuous in [0, 2] and differentiable in (0, 2)

So it satisfies Mean Value Theorem.

i.e., [f(b)f(a)](ba)=f(c)\frac{[ f(b) - f(a) ]}{ ( b-a)} = f'(c) ∀ c ∈ ( a, b)

f( 0) = 0

f(1) = 1

From Mean Value Theorem,

[f(1)f(0)](10)=f(c1)\frac{[ f(1) - f(0) ]}{ (1-0)} = f'(c1) ∀ c1 ∈ ( 0, 1)

(10)1=f(c1)\frac{(1-0)}{1} = f'(c1)

∴ f'(c1) = 1 ∀ c1 ∈ ( 0, 1)

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