Question

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 1

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

So it satisfies Mean Value Theorem.

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

f( 0) = 0

f(1) = 1

From Mean Value Theorem,

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

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

∴ f'(c1) = 1 ∀ c1 ∈ ( 0, 1)