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Let $a < c < b$. Suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,c)$ and on $(c,b)$, with $f′(x) > 0$ for all $x ∈ (a,c)$ and all $x ∈ (c,b)$. Show that $f$ is strictly increasing on $[a,b]$.

I can use the MVT to prove this, however I need to first show $f$ is differentiable at $c$. I do not know how to show this. Is there an alternative method?

kam
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  • Don't you need the IVT not MVT? – Andrew Li Mar 01 '18 at 02:02
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    If $f$ is continuous on $[a,c]$ and differentiable on $(a,c)$ with $f'(x) > 0$ for all $x \in (a,c)$ then show that $f$ is increasing on $[a,c]$. The same argument applies to $[c,b]$ and then, because $f$ is increasing on $[a,c]$ and $[c,b]$, it is increasing on $[a,b]$. – Trevor Gunn Mar 01 '18 at 03:25
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    $f$ need not be differentiable at $c.$ For example, join $(0,0), (1/2,1/2),(1,2)$ with line segments. – zhw. Mar 01 '18 at 03:40
  • See related question https://math.stackexchange.com/q/1845927/72031 – Paramanand Singh Mar 01 '18 at 09:27

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f is strictly increasing in $(a,c)$ as well as $(c,b)$. If we show that $f(x)<f(y)$ whenever $a<x<c$ and $c<y<b$ then it follows that f is strictly increasing on the whole of $(a,b)$. [ The case when $x=c$ or $y=c$ still remains but the argument below can be easily adapted to this case]. Pick points $\alpha$, $\beta$ such that $x<\alpha<c$ and $c<\beta <y$. Then $f(x)<f(\alpha) \leq f(c)$. [Indeed $f(\alpha)$ is less than the value of f at any point in $(\alpha,c)$ and we can use continuity to get $f(\alpha) \leq f(c)$]. Similarly, $f(c) < f(y)$, so $f(x)< f(c) < f(y)$. [ When $x=c$ we only need $\beta$ and when $y=c$ we only need $\alpha$].

Kavi Rama Murthy
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