# Darboux's theorem (analysis)

Darboux's theorem (analysis) In mathematics, Darboux's theorem is a theorem in real analysis, named after Jean Gaston Darboux. It states that every function that results from the differentiation of another function has the intermediate value property: the image of an interval is also an interval.

When ƒ is continuously differentiable (ƒ in C1([a,b])), this is a consequence of the intermediate value theorem. But even when ƒ′ is not continuous, Darboux's theorem places a severe restriction on what it can be.

Contents 1 Darboux's theorem 2 Proofs 3 Darboux function 4 Notes 5 External links Darboux's theorem Let {displaystyle I} be a closed interval, {displaystyle fcolon Ito mathbb {R} } be a real-valued differentiable function. Then {displaystyle f'} has the intermediate value property: If {displaystyle a} and {displaystyle b} are points in {displaystyle I} with {displaystyle ay>f'(b)} . Let {displaystyle varphi colon Ito mathbb {R} } such that {displaystyle varphi (t)=f(t)-yt} . If it is the case that {displaystyle f'(a)0} , we know {displaystyle varphi } cannot attain its maximum value at {displaystyle a} . (If it did, then {displaystyle (varphi (t)-varphi (a))/(t-a)leq 0} for all {displaystyle tin (a,b]} , which implies {displaystyle varphi '(a)leq 0} .) Likewise, because {displaystyle varphi '(b)=f'(b)-y<0} , we know {displaystyle varphi } cannot attain its maximum value at {displaystyle b} . Therefore, {displaystyle varphi } must attain its maximum value at some point {displaystyle xin (a,b)} . Hence, by Fermat's theorem, {displaystyle varphi '(x)=0} , i.e. {displaystyle f'(x)=y} . Proof 2. The second proof is based on combining the mean value theorem and the intermediate value theorem.[1][2] Define {displaystyle c={frac {1}{2}}(a+b)} . For {displaystyle aleq tleq c,} define {displaystyle alpha (t)=a} and {displaystyle beta (t)=2t-a} . And for {displaystyle cleq tleq b,} define {displaystyle alpha (t)=2t-b} and {displaystyle beta (t)=b} . Thus, for {displaystyle tin (a,b)} we have {displaystyle aleq alpha (t)

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