Why derivative of a closed real function is continous on open interval

[ManishR]

consider a real function $$f(x)$$ where $$x\in[a,b]$$
Why
$$f'(x),\; x\in(a,b)$$

 

[mathman]

f'(x) may not even exist, much less be continuous. You need much stronger assumptions on f(x) than just being defined on a closed interval.

 

[ManishR]

f'(x) may not even exist, much less be continuous. You need much stronger assumptions on f(x) than just being defined on a closed interval.

f'(x) does exist. forgot to mention in OP.

let f(x) = x^3 ?

 

[mathman]

Let f(x)=|x| for -1≤x≤1. f'(x) is discontinuous at 0.

 

[ManishR]

ok i think i may have misunderstood something.

let say f is continuous on the closed interval [a,b]

does it implies that f is differentiable on the open interval (a,b) .

 

[Office_Shredder]

No. f(x)=|x| on [-1,1] is continuous, but it's not differentiable at 0.

 

[ManishR]

No. f(x)=|x| on [-1,1] is continuous, but it's not differentiable at 0.

if f(x) range is [-1,1]
does
f'(x) exist at -1 and 1
or
f(x) is differentiable at -1 and 1.

 

[chiro]

mathman and shredder are right. You need stronger conditions that zeroth level continuity. For derivatives to exist for single variable real functions you need to prove that the left and right hand limits are equal for all values in the open set (a,b).

In analysis there are pretty rigorous definitions for this kind of thing. If you have started calculus, then showing that the appropriate limits from both sides should be sufficient

 

[RadioactivMan]

By definition the derivative of a function in a point c is

lim [f(c+h)-f(c)]/h
h->0

If the limit exist we can say that the derivative in point c is the result of the limit, but if a limit exists them the left and right hand limits are equal as chiro said.

If you are working on a [a,b] you can not define the derivative of f in a or b because of the definition of limit (epsilon-delta definition) since the function is not defined in a +- delta interval around a or b, but working of (a,b) is totally legal because exists a delta interval around any point in (a,b) where the function is defined .
In a the left hand limit does not exist because of definition them the derivative of f in a it is not defined.
The derivative function is a function that assigns to x the derivative of f in x, so for an interval [a,b] if we can get the derivative of f we get it at most just in the interval (a,b), so the derivative function is just defined in (a,b).

x^(1/2) is defined in x=0, but the derivative does not exist in x=0 because the limit does not exist, because an interval -δ<0<δ with δ>0, where the function is defined does not exist since x^(1/2) is not defined for x<0.

Everything I wrote is for single variable real functions.

 

[HallsofIvy]

if f(x) range is [-1,1]
does
f'(x) exist at -1 and 1
or
f(x) is differentiable at -1 and 1.

Are you talking about the "one-sided" derivatives? That is:
$$\displaytype \lim_{h\to 0^+} \frac{f(-1+h)- f(-1)}{h}$$
and
$$\displaytype \lim_{h\to 0^-} \frac{f(1+h)- f(1)}{h}$$