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LEARN MATH-TRICKS-The Notion OF IMVT

MATH-TRICK-5
This is a fundamental theorem of analysis which is an essential one for those attending an entrance exam in ISI, CMI.
INTERMEDIATE VALUE THEOREM:-
Let [a,b] is a closed and bounded interval and a function f : [a,b]->R be continuous on [a,b].
If f(a) ≠ f(b) then f attains every value between f(a) and f(b) at least once in the open interval (a,b).
Proof:- WLG, we assume f(a) < f(b).
Let k be a real number such that  f(a)<k<f(b).
Let us consider an another function g :[a,b]->R defined by g(x)=f(x)-k, where x is lying in [a,b].
Then g is continuous on [a,b],since f is continuous on [a,b].
g(a)=f(a)-k<0, g(b)=f(b)-k>0.
So, g(a) & g(b) are of opposite signs,by Bolzano's theorem there are at least one point c in (a,b) such that g(c)=0.
Therefore, f(c)=k.
The proof is complete.

Example. A function f :[0,1]->[0,1] is continuous on [0,1]. Prove that there exists a point c in [0,1],such that f(c)=c.
Sol:- If f(0)=0 or f(1)=1 the existance is proved.
We assume f(0)≠0 & f(1)≠1.
Let us consider a function g :[0,1]->R defined by g(x)=f(x)-x , x in [0,1] .
g is continuous on [0,1] & g(0)=f(0)>0 ,
since f(0) belongs to [0,1] & f(0)≠0.
Also g(1)=f(1)-1<0,since f(1) belongs to [0,1] & f(1)≠1 . By IVT there exists a point c in (0,1) such that g(c)=0. Therefore, f(c)=c.
Remark:- Here c is said to be a fixed point of the continuous map f .