Saved Bookmarks
| 1. |
State and prove Darboux theorem. |
|
Answer» Defineg(x)=f(x)−αxg(x)=f(x)−αx. Thenggis continuous and because[a,b][a,b]is compactggattains its minimum on[a,b][a,b]. Letxm∈[a,b]xm∈[a,b]be such thatg(xm)≤g(x)g(xm)≤g(x)for allx∈[a,b]x∈[a,b]. Ifxm∈(a,b)xm∈(a,b)theng′(xm)=0=f′(xm)−αg′(xm)=0=f′(xm)−αwhich shows the claim. 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). But this is a contradiction becausexmxmis the minimum. 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 |
|