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[parent] proof of Darboux's theorem (Proof)

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'$.

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 Fermat's theorem.



"proof of Darboux's theorem" is owned by paolini. [ full author list (2) | owner history (1) ]
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Cross-references: decreasing, increasing, continuous function, without loss of generality

This is version 3 of proof of Darboux's theorem, born on 2002-06-06, modified 2005-02-18.
Object id is 3056, canonical name is ProofOfDarbouxsTheorem.
Accessed 4967 times total.

Classification:
AMS MSC: 26A06 (Real functions :: Functions of one variable :: One-variable calculus)
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