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Grade 12Differential Calculus

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

Hemendu

10 Years agoGrade 12

4 Answers

Anoopam Mishra

10 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

Approved Tutor Answer10 Years ago

BY LMVT(Lagrange mean value theorem)

Apply theorem in at points (a,f(a)) & (b,f(b))

$\frac{f(a)-f(b)}{b-a} = f`(c)$ $a<c<b$

$\frac{b-a}{b-a} =1 = f`(c)$ 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

$\frac{f(\frac{a+b}{2}) -f(a)}{\frac{a+b}{2}-a} =f`(c_{1})$ $...............................(1)$ $a<c_{1}<\frac{a+b}{2}$

$\frac{f(b)-f(\frac{a+b}{2}) }{b-\frac{a+b}{2}} =f`(c_{2})$ $...............................(2)$ $\frac{a+b}{2}<c_{2}<b$

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

10 Years ago

My answer was exactly the same as of Akash. But you disapproved my answer.

Anoopam Mishra

10 Years ago

This is not fair!!!!!

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