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If the following data are given, prove that $f(a) \leq f(x) \leq f(b) $

f is differentiable on [a,b] and f'(x) $ \geq 0 \forall x \in (a,b) $

Is the following argument correct?

$f'(x) \geq 0 \implies f $ is increasing on (a,b) $ \implies f(a) \leq f(b) $

Let $x_0 \in (a,b) $

Since f is increasing $f(a) \leq f(x_0) \leq f(b) $ $$ \therefore \forall x \in (a,b) f(a) \leq f(x) \leq f(b) $$

S.Dan
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    You may want to use the mean value theorem to make things a bit more rigorous. – Cameron Williams Aug 11 '14 at 15:48
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    There is something missing: $f'(x) ≥ ?$ – miracle173 Aug 11 '14 at 15:48
  • $f$ can be differentiable on $(a,b)$. We need it to be continuous on $[a,b]$. Then you can use that argument, yes. – Darth Geek Aug 11 '14 at 15:52
  • f is differentiable on (a,b) $\implies $ f is continuous on (a,b) $\implies $ f is continuous on [a,b] right? – S.Dan Aug 11 '14 at 15:54
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    I'm not sure that the claim that a non-negative derivative implies the function to be increasing is actually easier than the original claim. Therefore (if you haven't already seen a proof of this) I'd suggest to prove this topological result. – Stefan Mesken Aug 11 '14 at 15:56
  • @Cameron How can I get the approach to use mean value theorum here? I don't know where to start. – S.Dan Aug 11 '14 at 15:56
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    @S.Dan Note that $f$ is continuous on $(a,b)$ doesn't imply that it is continuous on $[a,b]$. – Stefan Mesken Aug 11 '14 at 15:57
  • @S.Dan Consider: $$\frac{x^2-1}{\vert x^2-1\vert}$$

    It's continuous on $(-1,1)$ but not on $[-1,1]$

     – Darth Geek Aug 11 '14 at 16:02
  • Ok i got the brackets wrong, according to the interval given in the question, it is correct right? – S.Dan Aug 11 '14 at 16:11

2 Answers2

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Claim $f$ is continuous on $[a,b]$.

Proof $$\lim_{x\to a+}(f(x)-f(a))=\lim_{x\to a+} {f(x)-f(a)\over x-a} (x-a)=\lim_{x\to a+} {f(x)-f(a)\over x-a}\lim_{x\to a+}(x-a)=f'(a)\cdot0=0.$$ Do similarly at $b$.

Now suppose that $f(x)<f(a)$ for some $a\in(a,b)$. Use Mean Value Theorem on $[a,x]$, and we get a contradiction since $f'\ge0$ on $(a,x)$.

Do similarly when $f(x)>f(b)$.

Thus we get the result.

Jaehyeon Seo
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I think that the whole point of your problem is to prove that $\mathrm{f}'(x) > 0$ implies that $\mathrm{f}$ is increasing at $x$. Assume that $\mathrm{f}'(x) > 0$, by definition we have $$\lim_{h \to 0}\left(\frac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}\right) > 0$$ For sufficiently small, positive $h$ this implies that $$\frac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h} > 0$$ Multiplying through by $h$ gives $\mathrm{f}(x+h)-\mathrm{f}(x)>0$ and hence $\mathrm{f}(x+h)>\mathrm{f}(x)$.

Fly by Night
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  • This argument (correctly) shows $f$ is "increasing at $x$", but not that $f$ is increasing on some neighborhood of $x$. The latter is generally false without an additional hypothesis, e.g., $f'$ is continuous (and positive) at $x$, or (weaker) that $f'$ is strictly positive in a neighborhood of $x$. – Andrew D. Hwang Aug 11 '14 at 16:49
  • @user86418 You're right. Well spotted. – Fly by Night Aug 11 '14 at 17:16