Question

Let f(x) be defined on the open interval (0; 1). Suppose f is differentiable
and satisfies
f0(x) > 0
for all x in (0; 1). Does f have a (well-defined) inverse function? If yes, explain why. If
no, give a counterexample.

Answer 1

oops! Its f(x)>0 , not f0(x)>0

Answer 2

Do you mean f'(x) > 0 in the interval?

Answer 3

yes

Answer 4

So if f'(x) > 0 in the interval, that means that nowhere in this interval is f constant. For otherwise, f'(x) would be zero somewhere.

Agree?

Since f(x) is not constant anywhere on this interval and f(x) in increasing then for every value of x, there is a unique value of y. If this were not so, then at some point f'(x) would equal to zero, violating the condition of the problem that f(x) is increasing.

Agree?

So since for every x there is one and only one y , then for every y there is one and only one x.

Agree?

Since we can map y back to x by this logic, we have the inverse of y.

Answer 5

so does f has well defined inverse?

Answer 6

Yes, since it is unique by virtue of f(x) increasing and thus being a one-to-one correspondence between x and f(x)

Answer 7

ok thanks!