Answer #2 is not a good counterexample. That function doesn't have a derivative at x=0: f'(0)=lim{[f(0+h) -f(0)]/h} = lim f(h)/h = 0 if h<0 and =1 if h>0 so the limit is undefined. So f is not differentiable at x=0.
The wikepdia site (way at the end) says it is true. I couldn't find any proofs online too easily. The classic differentiable but with a discontinuous derivative example is x^2*sin(1/x) but (I guess) that function does satisfy the intermediate value property, even though it is discontinuous.
Here ya go! It is called Darboux's theorem and the proof is:
Without loss of generality we might and shall assume f' + (a) > t > f' − (b). Let g(x) := f(x) - tx. Then g'(x) = f'(x) − t, g' + (a) > 0 > g' − (b), and we wish to find a zero of g'.
Since g is a continuous function on [a,b], by the extreme value theorem it attains a maximum on [a,b]. This maximum cannot be at a, since g' + (a) > 0 so g is locally increasing at a. Similarly, g' − (b) < 0, so g is locally decreasing at b and cannot have a maximum at b. So the maximum is attained at some c in (a,b). But then g'(c) = 0 by Fermat's theorem (stationary points).
In addition to being differntiable on an open interval, f must also be continuous on a closed interval. Otherwise you can find counterexamples such as the one above.
Given continity and differntiability, this follows from the completeness property of real numbers.
Complex numbers have a similar completeness property.
Comments
Answer #2 is not a good counterexample. That function doesn't have a derivative at x=0: f'(0)=lim{[f(0+h) -f(0)]/h} = lim f(h)/h = 0 if h<0 and =1 if h>0 so the limit is undefined. So f is not differentiable at x=0.
The wikepdia site (way at the end) says it is true. I couldn't find any proofs online too easily. The classic differentiable but with a discontinuous derivative example is x^2*sin(1/x) but (I guess) that function does satisfy the intermediate value property, even though it is discontinuous.
Here ya go! It is called Darboux's theorem and the proof is:
Without loss of generality we might and shall assume f' + (a) > t > f' − (b). Let g(x) := f(x) - tx. Then g'(x) = f'(x) − t, g' + (a) > 0 > g' − (b), and we wish to find a zero of g'.
Since g is a continuous function on [a,b], by the extreme value theorem it attains a maximum on [a,b]. This maximum cannot be at a, since g' + (a) > 0 so g is locally increasing at a. Similarly, g' − (b) < 0, so g is locally decreasing at b and cannot have a maximum at b. So the maximum is attained at some c in (a,b). But then g'(c) = 0 by Fermat's theorem (stationary points).
found at http://en.wikipedia.org/wiki/Darboux%27s_theorem_%...
Not enough information.
In addition to being differntiable on an open interval, f must also be continuous on a closed interval. Otherwise you can find counterexamples such as the one above.
Given continity and differntiability, this follows from the completeness property of real numbers.
Complex numbers have a similar completeness property.
http://en.wikipedia.org/wiki/Complete_space
I think this is false.
Look at f(x) = 0 if x <=0,
f(x) = x >= 0.
The derivative is the unit step function, which
is not continuous at 0 and there is no value
of x for which f'(x) = 0.
Hope this makes sense!
Do you really need to prove it, or just understand it? In either case, information and the proof can be found on the Wikipedia page.