Is a positive derivative sufficient to show ln is unbounded near zero?

[Eclair_de_XII]

Homework Statement

Define $f:(-1,0)\rightarrow \mathbb{R}$ by $f(x)=-\ln(-x)$. Show that $f$ is unbounded.

Relevant Equations

A function $f$ is said to be unbounded if for all positive numbers $M$, there is a $y$ in $\textrm{dom}(f)$ such that $|f(y)|>M$.

So far, I found the derivative of $f$:

\begin{align*}
\frac{d}{dx}\,f(x)&=&-\frac{d}{dx}\,\ln(-x)\\
&=&-\left(\frac{1}{(-x)}\right)(-1)\\
&=&-\frac{1}{x}
\end{align*}

$f'(x)$ is always positive and never zero on its domain.

Hence, $f$ does not have a local maximum and is always increasing on the interval $(-1,0)$.

Are these conditions sufficient to argue that $\ln$ is unbounded near zero?

 

[Office_Shredder]

What about f(x)=x? The derivative is always positive as well. Is that unbounded?

 

[FactChecker]

That is not what they are looking for. You should directly use the definition of "unbounded" that you gave. Assume that you have a value for $M \gt 0$ and determine a region $R=(0,r]$ for which $x\in R$ implies $ln(x)\gt M$.

EDIT: The above is wrong. Determine a value, $r \in (-1,0)$ where $|f(r)|\gt M$.

 

[Eclair_de_XII]

It's not as simple as setting $f(r) = M + 1$ and then solving for $r$, is it?

$r=-\exp[-(M+1)]$

\begin{align*}
f(r)&=&-\ln(-r)\\
&=&-\ln[-(-\exp[-(M+1)])]\\
&=&-\ln[\exp[-(M+1)]]\\
&=&\ln[\exp(M+1)]\\
&=&M+1\\
&>&M
\end{align*}

 

[Office_Shredder]

Is r inside the interval you are supposed to be working on?

 

[Eclair_de_XII]

Let me see:

1. The $\exp$ function is always positive, so $-\exp(x)$ for some $x$ is always negative.
2. $e>2>0$. $M$ is positive, so the sum $M+1$ is positive. $|e^{-(M+1)}|=\frac{1}{e^{M+1}}<1$, as a result.

I should think that $r=-\exp[-(M+1)]$ is in $(-1,0)$

 

[Office_Shredder]

Looks right to me.

 

[Eclair_de_XII]

Arr, thanks for the help and giving that counter-example that I overlooked.

 

[Mark44]

Define $f:(-1,0)\rightarrow \mathbb{R}$ by $f(x)=-\ln(-x)$. Show that $f$ is unbounded.

It would have been simpler to work with $g(x) = \ln(x)$ on the interval (0, 1). It would have made the arithmetic a bit less tedious. This function is the reflection across both the x and y axes of the one you have. Both f and g are unbounded for x near 0.