Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). Suppose that f(a) = aand f(b) = b. Show that there is c ∈ (a, b) such that f′(c) = 1. Further, show that thereare distinct c1, c2 ∈ (a, b) such that f′(c1) + f′(c2) = 2.

Anoopam Mishra
126 Points
8 years ago
Using Lagrange mean value theorem we know that $f'(c)=(f(b)-f(a))\div (b-a)$ for atleast one $c\in \left [ a,b \right ]$. You can see easily that $f'(c)=1$.
Now the second part, take $y = (a+b)/2$. We will apply LMVT for two intervals $\left [ a,a+b/2 \right ] and \left [ a+b/2,b \right ]$. There exists atleast one $c1\in \left [ a,a+b/2 \right ] and\ c2\in \left [ a+b/2,b \right ]$ for which $f'(c1)=f(y)-f(a)/(y-a) = f(y)-a/(a+b)/2-a = 2f(y)-2a/b-a\ and\ f'(c2)=f(b)-f(y)/b-y = b-f(y)/b-(a+b)/2 = 2b-2f(y)/b-a.$
It is easy to see that $f'(c1)+f'(c2) = 2$.
Aakash
110 Points
8 years ago
BY LMVT(Lagrange mean value theorem)
Apply theorem in at points (a,f(a)) & (b,f(b))

$\frac{f(a)-f(b)}{b-a} = f'(c)$  $a
$\frac{b-a}{b-a} =1 = f'(c)$ This prove the first part that there exist a point c between a & b such that $f'(c) = 1$

SECOND PART

now break the interval (a,b) in two Parts in $( a, \right\frac{a+b}{2} )$ $&$$\left (\frac{a+b}{2}, \right b )$

now apply LMVT indivisually in both interval

$\frac{f(\frac{a+b}{2}) -f(a)}{\frac{a+b}{2}-a} =f'(c_{1})$  $...............................(1)$ $a
$\frac{f(b)-f(\frac{a+b}{2}) }{b-\frac{a+b}{2}} =f'(c_{2})$ $...............................(2)$$\frac{a+b}{2}
Adding $(1)$ and $(2)$
also put $f(a) =a$ $And$ $f(b)=b$
you got

$f'(c_{1}) + f'(c_{2}) =2$

This proves second part also that there exist c1 & c2 between (a,b) such that f’(c1)+f’(c2) =2

$PLEASE$ $APPROVE$
Anoopam Mishra
126 Points
8 years ago
My answer was exactly the same as of Akash. But you disapproved my answer.

Anoopam Mishra
126 Points
8 years ago
This is not fair!!!!!