Suppose that $f$ is one-to-one and continuous on [$a,b$]. Prove that $f$ is either strictly increasing or strictly decreasing on [$a,b$].
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$\begingroup$ Suppose it weren't. Use the intermediate value theorem to derive a contradiction. $\endgroup$ – Daniel Fischer♦ Oct 23 '13 at 22:30
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Hint Use the intermediate-value theorem to get a contradiction if you assume $$\exists x,y \in [a,b] : f'(x) < 0 \wedge f'(y) > 0$$ Use Weierstraß to acqurie a differentiable function with norm arbitrarily close to $f$.
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$\begingroup$ Most (in the sense that the complement is of the first category) elements of $C^0([a,b])$ are nowhere differentiable. $\endgroup$ – Daniel Fischer♦ Oct 23 '13 at 22:27
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$\begingroup$ @All sorry, updated argument. Weierstraß works here. $\endgroup$ – AlexR Oct 23 '13 at 22:32
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Hint: Go by contradiction. If its not monotone then we have some $p \le q \le r$ such that
$$f(p) \ge f(q) \le f(r)$$
Then apply the intermediate value theorem to get a contradiction.
