I am given the following function:
$$f: \mathbb{R} \setminus \{ 1 \} \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{\sqrt{x^2 + a^2}}{x - 1}$$
with $a \in \mathbb{R}^*$. I have to check if the function is monotone on the interval $(0, \infty) \setminus \{ 1 \}$. Firstly, I found the derivative of the function:
$$f'(x) = - \dfrac{x + a^2}{(x-1)^2\sqrt{x^2+a^2}} \hspace{2cm} \forall \hspace{.1cm} x \in \mathbb{R} \setminus \{ 1 \}$$
In order to find the monotony of the function I tried to find the values of $x$ for $f'(x) = 0$. I believe this is where I made a mistake, or so I'm told. This is what I did:
$$f'(x) = 0 \Rightarrow -x-a^2 = 0 \Rightarrow x=-a^2$$
So we have a local extremum point in $x = -a^2$.
Then, I found the derivative on different intervals to see where the function is monotonic. On the inteval $(-\infty, -a^2)$ I found that the derivative is positive, on the interval $(-a^2, 1)$ I found that the derivative is negative and on the interval $(1, \infty)$, again, I found that the derivative is negative.
So, on the interval $(-a^2, \infty) \setminus \{1\}$ the derivative is negative, so the function is monotonic. Since $a \in \mathbb{R}^*$, we have that $-a^2 < 0$. So, given that the function is monotonic on $(-a^2, \infty) \setminus \{1\}$ and $-a^2 < 0$, we can conclude that the function is monotonic on $(0, \infty) \setminus \{ 1 \}$.
The problem is that my textbook disagrees with me. The actual answer is that the function is NOT monotonic on the interval $(0, \infty) \setminus \{ 1\}$. Explanations are not given.
Someone told me that my mistake is in solving $f'(x) = 0$, saying that $-x = a^2$ has no solutions since something squared cannot be equal to something negative, therefore, we don't have any $x$, where $f'(x) =0$.
But I don't understand this correction. I mean, if we have that $x = -5$, we would have $-(-5) = a^2$. In other words, $5 = a^2$, which certainly is possible.
So what is wrong? Why is the function NOT monotonic on $(0, \infty) \setminus \{ 1 \}$. If you could explain this in detail, so I can understand it fully, it would be perfect.
You can say that $f(x)$ is piecewise-monotonic on $(0,\infty)\setminus\{1\}$ because it's monotonically decreasing on each of the intervals $(0,1)$ and $(1,\infty)$. However, the problem with saying it's monotonic on that set is the following. There is a vertical asymptote at $x=1$: $\lim_{x\to 1^-}f(x)=-\infty$ and $\lim_{x\to 1^+}f(x)=+\infty$. This means that we can't say $f(x)$ is decreasing (and hence monotonic) on $(0,\infty)\setminus\{1\}$. In fact, we have $f(0.5)<0<f(1.5)$ for all $a$, which contradicts the idea that $f(x)$ is decreasing on $(0,\infty)\setminus\{1\}$. More generally, $f(x)$ is negative on $(0,1)$ and positive on $(1,\infty)$.
A similar question to ask is whether $g(x)=\frac 1x$ is monotonic on $(-\infty,\infty)\setminus\{0\}$. It's not because of the asymptote at $x=0$.