# Math Help - continious function

1. ## continious function

A function f: A-> R is increasing if f(x)<= f(y) for every x, y in A such that x<=y.

suppose that f:[a,b] ->R is increasing and that for every Y in [f(a), f(b)] there is some x[0] in [a,b] sucht that f( x[0] ) = Y

How do i show that fis a continious function?

2. I have an idea.

The limit $\lim_{x\to c^-}f(x)=\inf \{f(x)|x
And $\lim_{x\to c^+}f(x)=\sup \{f(x)|x>c\}=f(c)$
So the left and right limits exists and are equal to the functional evaluation.

Because since the function is increasing it places the role of the least upper bound.

3. ## increasing functions part 2

Thankz...
i want to show that the asuumption that f is increasing is essential by
finding an example of a function f:[-1,1] ->R which has the property that for every Y in [f(-1), f(1)] there is some x[0] in [-1,1] such that f( x[0] ) = Y , bu it is not continious and so fcan't be increasing.

Does any one know any examples with these values for a=-1 and b=1?

thankz

4. Originally Posted by dopi
Thankz...
i want to show that the asuumption that f is increasing is essential by
finding an example of a function f:[-1,1] ->R which has the property that for every Y in [f(-1), f(1)] there is some x[0] in [-1,1] such that f( x[0] ) = Y , bu it is not continious and so fcan't be increasing.
What you actually are asking is for the converse of the intermediate value theorem. You know that any continous function on a closed interval satisfies the Intermediate property. However you wish to now if a functions satisfies the Intermediate property then it is not necessarly continous.

Consider a semicircle $f:[-1,1]\to \sqrt{1-x^2}$
Now take any point not on endpoints of its extreme value and move it done "slightly" IVT still works but the function has replacable discontinuity.