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$f: (a,b) \to \Bbb{R}$ is differentiable and $f'(x)>0$ at all points but at $c$ where $f'(c) = 0$.

I need to prove that $f$ is strictly increasing.

I thought to split the intervals to $(a,c)$ and $(c,b)$ and use the continuity of $f$ at $c$, but I'm not sure how to explain that.

More generally, I understand that this is true for a finite number of critical points, how do I explain that too?

Help please

Domates
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Itay4
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    Sorry if I wasn't clear. The first case is for one critical point, and the second is for a finite number of critcal points. – Itay4 Jan 10 '17 at 20:52

3 Answers3

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Hint: You can show that $f(x)<f(c)<f(y)$, when $x<c<y$ using contradiction and mean value theorem. This is sufficient to prove that $f$ is strictly increasing everywhere.

For finite number of $\{c_i\}_{i\in I}$, $I = \{1,\ldots,n\}$ such that $a = c_0 < c_1<c_2<\ldots<c_n < c_{n+1} = b$ and $f'(c_i)=0,\ i\in I$, $f'(x)>0$ when $x\neq c_i,\ i\in I$, consider invervals $A_i=(c_{i-1},c_{i+1}),\ i\in I$. By the result for one point, $f$ is strictly increasing on each $A_i$ and since they cover $(a,b)$, it is strictly increasing everywhere.

Ennar
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Yes, it can be done for a finite number of dicontinuities: take $x<y$ and pick $z\in (x,y)$ such that no point in $(x,z)$ has derivative $0$.

You can prove each of the following inequalities with the mean value theorem:

$f(x)<f(z)\leq f(y)$.

Asinomás
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Let $x_1<x_2\in \left( a,b\right)$.

First let's prove under the assumption that $f^\prime > 0$ on $\left( a,b \right)$.

Hint 1

Lagrange's mean value theorem tells us there is an $x\in \left( x_1, x_2 \right)$ such that $$f^{\prime}\left(x\right)=\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}$$

Hint 2

$f^{\prime}\left(x\right)>0$ so we have $f\left(x_{2}\right)-f\left(x_{1}\right)>0$

So $f$ is strictly monotonic increasing on $\left( a,b \right)$.

Let's now weaken our assumption to the original one presented. Examining the cases where $x_1 < c < x_2$ and $\left(x_{1}<x_{2}\leq c\right)\vee\left(c\leq x_{1}<x_{2}\right)$ seperately sounds like a good idea, and we should be able to use the above result to prove them quite easily!

Can you see how to proceed from here?

As for a finite number of critical points: I'd try induction on the number of critical points.

user401346
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