**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 .