If f SubsetEqual Ropf is an interval and f I rightarrow Ropf

If f SubsetEqual Ropf is an interval and f: I rightarrow Ropf, then we say f is strictly increasing on I if f(x)

Solution

Prove the contrapositive instead: if f is not strictly increasing and not strictly decreasing, then it is not-one-ot-one.

there are points a<b<c such that f(a)< f(b) and f(b) > f(c). Either f(a)=f(c)
(in which case f is not one-to-one), or f(a)<f(c), or f(c)<f(a). for all a , b and c belongs to I.

If f(a)<f(c)<f(b), then by the Intermediate Value Theorem there exists d (a,b) such that f(d) = f(c);

hence f is not one-to-one.

b) Given that f is injective , and g ; f(I) ---> I is defined as g = f^-1(I)

We have to show that is is continus f is one to one in to is\'t range which is the intervel ( f(a) , f(b) )

and   f is strictly increasing or strctly decreasing therefor g = f^-1(I) is also strictly increasing or strictly decresing . and so it is continous.