Answer to Question #191286 in Real Analysis for Parul

Question

Suppose that f :[0, 2] ->R is continuous on [0, 2] and differentiable on (0, 2), and that f (0) = 0, f (1) =1, f (2) =1. (i) Show that there exists c1 belongs to(0,1) such that f'(c1) =1 (ii) Show that there exists c2 belong to (1,2)such that f'(c2)=0(iii) Show that there exists c belongs to(0, 2) such that f'=1/3

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., [ f(b) - f(a) ] / ( b-a) = f'(c) ∀ c ∈ ( a, b)

i) f( 0) = 0

f(1) = 1

From Mean Value Theorem,

[ f(1) - f(0) ] / (1-0) = f'(c1)    ∀ c1 ∈ ( 0, 1)

(1-0)/1 = f'(c1)

∴ f'(c1) = 1     ∀ c1 ∈ ( 0, 1)

ii)  f( 1) = 1

f(2) = 1

From Mean Value Theorem,

[ f(2) - f(1) ] / (2-1) = f'(c2)   ∀ c2 ∈ ( 1, 2)

(1-1)/1 = f'(c2)

∴ f'(c2) = 0    ∀ c1 ∈ ( 2, 1)

iii)  f( 0) = 0

f(2) = 1

From Mean Value Theorem,

[ f(2) - f(0) ] / (2-0) = f'(c)   ∀ c ∈ ( 0, 2)

(1-0)/2 = f'(c)

∴ f'(c) = 1/2    ∀ c1 ∈ ( 0, 2)