How to deduce the monotonicity of a function from another function?

Question

I have the following task and I am a bit confused by the solution of it. In the official solution it's stated that only the first assumption is correct. But in my approach I am always coming to the end that the first and the third assumptions are correct. I would be happy if someone could verify if my solution is correct and there is a mistake in the solutions or to explain where I am thinking wrong.

Task

The function $f:\mathbb{R}\rightarrow\mathbb{R}\setminus\{0\}$ is a strict monotonic growing function. Let $g:\mathbb{R}\rightarrow\mathbb{R}$, $g(x)=\frac{1}{f(x)}$.

Which of the following is correct?

1. If $f$ doesn't take on positive function values, $g$ is strictly monotonically decreasing.
2. If $f$ doesn't take on positive function values, $g$ is strictly monotonically increasing.
3. If $f$ only takes on function values which are bigger than $-1$, $g$ is strictly monotonically decreasing.
4. If $f$ only takes on function values which are less than $1$, $g$ is strictly monotonically increasing.

My Solution

First we check what happens when $f$ doesn't take on positive function values. That means that for every $x\in\mathbb{R}$, $f(x)<0$ holds. We know that if we have $x,y\in\mathbb{R}$ with $x<y$ then $f(x)<f(y)$. In this case the function $f$ looks like this $f:\mathbb{R}\rightarrow\mathbb(-\infty,0)$. Where $(-\infty,0)$ is an interval. Then let $h$ be the following function $h:(-\infty,0)\rightarrow\mathbb{R}$, $h(x)=\frac{1}{x}$. We can see that $h$ is strictly monotonic decreasing. If we have $x,y\in(-\infty,0)$ with $x<y$ then we can see that $f(x)>f(y)$ holds. Now the following holds: $(h\circ f)(x)=h(f(x))=\frac{1}{f(x)}=g(x)$. Because the function composition of one strictly monotonic increasing and one strictly decreasing function is strictly decreasing again, we can say that the first assumption is a correct one and the second one is wrong. From this reasoning we can also deduce that the fourth assumption is wrong.

Now we are looking at the third assumption. Where we look at the case where $f(x)>0$. Here our function $f$ looks like the following $f:\mathbb{R}\rightarrow(0,\infty)$. Then we let $p$ be the following function $p:(0,\infty)\rightarrow\mathbb{R}$, $p(x)=\frac{1}{x}$. If we have $x,y\in(0,\infty)$ with $x<y$ we have $p(x)>p(y)$ and thus we have a monotonic decreasing function again. Now the following holds again $(p\circ f)(x)=p(f(x))=\frac{1}{f(x)}=g(x)$. And because the function composition of one strictly monotonic increasing and one strictly decreasing function is strictly decreasing again. Thus the third assumption is correct as well.

So where is my error, or is the solution incorrect?

Answer 1

[Edit: Fixing example of $f$ to comply with assumption $f(x)>-1$.]

The problem with your reasoning is you only looked at the case where $f(x)>0$. In the third assumption, $f$ may take on values that are negative and then have a jump discontinuity and take on values that are positive. For example, we could have $f(x)=-\frac{x}{x-1}$ for $x<0$ and $1+x$ for $x\geq 0$. Then $g(x)=\frac{1}{x}-1<0$ for $x<0$ and $\frac{1}{1+x}>0$ for $x\geq 0$, and so it fails to be decreasing across $x=0$.

Basically the problem is that $x\mapsto \frac{1}{x}$ is not monotone on domains with positive and negative values. And since $f$ need not be continuous, its range (which becomes the domain for $x\mapsto \frac{1}{x}$) can be disconnected, allowing it to include positive and negative values even though it doesn't include $0$.