- #1
Chsoviz0716
- 13
- 0
Hi, I have a question about the definition of derivative.
As far as I know, for a real valued function f defined on a subset of R, the derivative of f at a is
(f(x+h)-f(x))/h as h → 0.
And if it exists it's said f is differentiable at x.
What if I define f : Q → R as follows,
f(x) = sin(x) (the domain of f is all rational numbers.)
Then it satisfies every condition to be a differentiable function, thus it's a differentiable function on its domain.
Is this a valid argument?
Or what about f : [0,1]U[2,3] → R defined as follows,
f(x) = 1 when x belongs in the first closed interval
2 when x belongs in the second closed interval
f'(x) at points 1 and 2 seem to exist. But do they really?
And one small question; when you have f'(x) = ∞, it implies f isn't defined at x. Is it even possible to have such an equality? If f(x) isn't defined, how can f'(x) can be calculated in the first place when the definition of f'(x) requires the existence of f(x)?
Thank you in advance.
As far as I know, for a real valued function f defined on a subset of R, the derivative of f at a is
(f(x+h)-f(x))/h as h → 0.
And if it exists it's said f is differentiable at x.
What if I define f : Q → R as follows,
f(x) = sin(x) (the domain of f is all rational numbers.)
Then it satisfies every condition to be a differentiable function, thus it's a differentiable function on its domain.
Is this a valid argument?
Or what about f : [0,1]U[2,3] → R defined as follows,
f(x) = 1 when x belongs in the first closed interval
2 when x belongs in the second closed interval
f'(x) at points 1 and 2 seem to exist. But do they really?
And one small question; when you have f'(x) = ∞, it implies f isn't defined at x. Is it even possible to have such an equality? If f(x) isn't defined, how can f'(x) can be calculated in the first place when the definition of f'(x) requires the existence of f(x)?
Thank you in advance.