# another darboux question

• Feb 2nd 2010, 01:29 AM
derek walcott
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).

• Feb 2nd 2010, 03:17 AM
tonio
Quote:

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).

$\displaystyle f(x)=|x|$ isn't derivable at $\displaystyle x=0$ and thus there's no contradiction to anything.

Tonio
• Feb 2nd 2010, 12:29 PM
Drexel28
Quote:

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).

As tonio said Darboux's theorem says that if $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and differentiable on $\displaystyle (a,b)$ then for any $\displaystyle \min\left\{f'_{-}(a),f'_{+}(b)\right\}<k<\max\left\{f'_{-}(a),f'_{+}(b)\right\}$ there exists some $\displaystyle c\in(a,b)$ such that $\displaystyle f'(c)=k$. Thus, Darboux's theorem is not violated here since $\displaystyle f$ is not differentiable on ANY open ball of $\displaystyle 0$. So there is no way to even apply it.