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 ;
F(a)<0 and f(b)>0 .
Science f(x) is continuous and f(a)<0 , therefore f(x) will be negative in the neighborhood of a .
Again Since f(x) is continuous and f(b)>0 therefore f(x) will be positive in the neighborhood of b .
The set of values of x between a and b which make f(x) positive is bounded below by a and hence possesses an exact lower bound k .
Hence ; a<k<b
In this way we find that f(x) is positive in the interval
k<x<b and is negative or zero in the interval
a is less then and equal to x<k .
Since f(x) is continuous at x=k , therefore by the definition of continuity,
f(k-0)=f(k)=f(k+0) .
Since f(x) is negative or zero in
a is less then and equal to x<k , therefore f(k-0) must be negative and zero , therefore f(k) which is equal to f(k-0) must be negative or zero .
We shall now show that f(k) can not be negative .
Since f(x) is positive in the interval k<x<b , that is ;
b>x>k , therefore f(k+0) can not be negative and since ;
f(k)=f(k+0) , therefore f(k) can not be negative
Hence it follows that f(k)=0 and the Theorem is therefore proved .
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