- #1
lolgarithms
- 120
- 0
Is a bijective continuous function:[a,b]->[f(a),f(b)] differentiable?
I think it has to be.
continuity between two distinct values of f(a) and f(b): it got to take all the values between f(a) and f(b) at x in [a,b], by the intermediate value theorem.
if f is bijective, at [a,b], f(x) can't go up and then down at [a,b]. it has to be monotonically increasing for it to be bijective.
so a function can only be non-differentiable at a set of points of measure zero. (like the vertical tangent of f(x)=x^1/3 at 0)
I think it has to be.
continuity between two distinct values of f(a) and f(b): it got to take all the values between f(a) and f(b) at x in [a,b], by the intermediate value theorem.
if f is bijective, at [a,b], f(x) can't go up and then down at [a,b]. it has to be monotonically increasing for it to be bijective.
so a function can only be non-differentiable at a set of points of measure zero. (like the vertical tangent of f(x)=x^1/3 at 0)