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Revision difference : proof of Darboux's theorem |
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Without loss of generality we migth and shall assume $f'_{+}(a)>t>f'_{-}(b)$. Let $g(x):=f(x)-tx$. Then $g'(x)=f'(x)-t$, $g'_{+}(a)>0>g'_{-}(b)$, and we wish to find a zero of $g'$.
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WLOG, assume $f'_{+}(a)>t>f'_{-}(b)$. Let $g(x)=f(x)-tx$. Then $g'(x)=f'(x)-t$, $g'_{+}(a)>0>g'_{-}(b)$, and we wish to find a zero of $g'$.
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Since $g$ is a continuous function on $[a,b]$, it attains a maximum on $[a,b]$. This maximum cannot be at $a$, since $g'_{+}(a)>0$ so $g$ is locally increasing at $a$. Similarly, $g'_{-}(b)<0$, so $g$ is locally decreasing at $b$ and cannot have a maximum at $b$. So the maximum is attained at some $c \in (a,b)$. But then $g'(c)=0$ by \PMlinkname{Fermat's theorem}{FermatsTheoremStationaryPoints}.
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$g$ is a continuous function on $[a,b]$, so it attains a maximum on $[a,b]$. This maximum cannot be at $a$, since $g'_{+}(a)>0$ so $g$ is locally increasing at $a$. Similarly, $g'_{-}(b)<0$, so $g$ is locally decreasing at $b$ and cannot have a maximum at $b$. So the maximum is attained at some $c \in (a,b)$. But then $g'(c)=0$ by \PMlinkname{Fermat's theorem}{FermatsTheoremStationaryPoints}.
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