Answer to Question #314400 in Real Analysis for Pankaj

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 c↓1∈ (0,1)such that f'(c↓1) =1

(ii) Show that there exists c↓2 ∈ (1,2)such that f'(c↓2) =0.

(iii) Show that there exists c ∈ (0,2)such that f'(c) =1/3

Answer 1

$\left( i \right) Lagrange‘s\,\,theorem\,\,on\,\,\left[ 0,1 \right] :\\f\left( 1 \right) -f\left( 0 \right) =f'\left( \xi \right) \left( 1-0 \right) \Rightarrow f'\left( \xi \right) =1-0=1\\\left( ii \right) Lagrange’s\,\,theorem\,\,on\,\,\left[ 1,2 \right] :\\f\left( 2 \right) -f\left( 1 \right) =f'\left( \xi \right) \left( 2-1 \right) \Rightarrow f'\left( \xi \right) =1-1=0\\\left( iii \right) f'\in C\left[ 0,2 \right] ,f'\left( \xi _1 \right) =0,f'\left( \xi _2 \right) =1\Rightarrow \\\Rightarrow by\,\,IMVT\,\,there\,\,exists\,\,\xi \,\,between\,\,\xi _1,\xi _2: f'\left( \xi \right) =\frac{1}{3}$