InterviewSolution
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InterviewSolution
| 1. |
State and prove Darboux theorem. |
| Answer» <p>Defineg(x)=f(x)−αxg(x)=f(x)−αx.</p><p> Thenggis continuous and because[a,b][a,b]is compactggattains its minimum on[a,b][a,b]. </p><p>Letxm∈[a,b]xm∈[a,b]be such thatg(xm)≤g(x)g(xm)≤g(x)for allx∈[a,b]x∈[a,b].</p><p> Ifxm∈(a,b)xm∈(a,b)theng′(xm)=0=f′(xm)−αg′(xm)=0=f′(xm)−αwhich shows the claim.</p><p>Ifxm=axm=atheng′(xm)=g′(a)<0g′(xm)=g′(a)<0. Becauseggis continuous andg′(a)<0g′(a)<0there existsδ>0δ>0such that ifx∈(a,a+δ)x∈(a,a+δ)theng(x)<g(a)=g(xm)g(x)<g(a)=g(xm). </p><p>But this is a contradiction becausexmxmis the minimum.</p><p> Ifxm=bxm=bthen again there isδ>0δ>0such that ifx∈(b−δ,b)x∈(b−δ,b)theng(x)<g(b)=g(xm)g(x)<g(b)=g(xm)becauseg</p> | |