Thread: another darboux question

Let K be a D-domain and let f:K--R be differentiable on the interval [a,b] as a subset of K (a< b).


f(x) = { 1 if x>0
{ -1 if x<0
is the derivate of f(x) = |x|.
Why doesn't this contradict Darboux's theorem?

I don't understand how this DOESN'T contradict Darboux. There is discontinuity in the f'(x) graph because it jumps from -1 to 1. Also it violates the IVT because there is no u for which f(a)<u<f(b).

Please help

Originally Posted by derek walcott

Let K be a D-domain and let f:K--R be differentiable on the interval [a,b] as a subset of K (a< b).


f(x) = { 1 if x>0
{ -1 if x<0
is the derivate of f(x) = |x|.
Why doesn't this contradict Darboux's theorem?

I don't understand how this DOESN'T contradict Darboux. There is discontinuity in the f'(x) graph because it jumps from -1 to 1. Also it violates the IVT because there is no u for which f(a)<u<f(b).

Please help

isn't derivable at and thus there's no contradiction to anything.

Tonio

Originally Posted by derek walcott

Let K be a D-domain and let f:K--R be differentiable on the interval [a,b] as a subset of K (a< b).


f(x) = { 1 if x>0
{ -1 if x<0
is the derivate of f(x) = |x|.
Why doesn't this contradict Darboux's theorem?

I don't understand how this DOESN'T contradict Darboux. There is discontinuity in the f'(x) graph because it jumps from -1 to 1. Also it violates the IVT because there is no u for which f(a)<u<f(b).

Please help

As tonio said Darboux's theorem says that if is continuous on and differentiable on then for any there exists some such that . Thus, Darboux's theorem is not violated here since is not differentiable on ANY open ball of . So there is no way to even apply it.