**Theorem :-****A continuous function which has opposite signs at two points vanishes at least once between these points , that is if**

*f(x)*be continuous in the closed interval*[a,b]*and*f(a)*and*f(b)*have opposite signs , then there is at least one value of*x*between*a*and*b*for which*f(x)=0 .*

__Proof :-__For the sake of definiteness , let us suppose that ;

*and*

**F(a)<0**

**f(b)>0 .**Science

*is continuous and*

**f(x)***, therefore*

**f(a)<0***will be negative in the neighborhood of*

**f(x)***.*

**a**Again Since

*is continuous and*

**f(x)***therefore*

**f(b)>0***will be positive in the neighborhood of*

**f(x)***.*

**b**The set of values of

*between*

**x***and*

**a***which make*

**b***positive is bounded below by a and hence possesses an exact lower bound*

**f(x)**

**k .**Hence ;

**a<k<b**In this way we find that

*is positive in the interval*

**f(x)***and is negative or zero in the interval*

**k<x<b**

**a is less then and equal to x<k .**Since

*is continuous at*

**f(x)***, therefore by the definition of continuity,*

**x=k**

**f(k-0)=f(k)=f(k+0) .**Since

*is negative or zero in*

**f(x)***therefore*

**a is less then and equal to****x<k ,***must be negative and zero , therefore*

**f(k-0)***which is equal to*

**f(k)***must be negative or zero .*

**f(k-0)**We shall now show that

*can not be negative .*

**f(k)**Since

*is positive in the interval*

**f(x)***, that is ;*

**k<x<b***, therefore*

**b>x>k***can not be negative and since ;*

**f(k+0)***, therefore*

**f(k)=f(k+0)***can not be negative*

**f(k)**Hence it follows that

*and the Theorem is therefore proved .*

**f(k)=0**