How do I define my function? I have this function here.
$$f(x)=\frac{\frac{x-14}{x-2}-1}{7+ \frac{4}{x-2} }$$
I can see that when $x=2$ and when $$x=\frac{10}{7}$$
it's undefined.
But when I simplify this into this:$$ f(x)=\frac{-12}{7x-10}$$
$x=10/7$ is still not defined, however, $x=2$ is defined.
So my question is, what does this mean? Is the original expression defined for $x=2$ or not? Or is there something else I'm missing here?
I'm wondering this because my teacher asked me to calculate $f(x)=-3$ and then double-check it by plugging it into the original expression. When I calculate $f(x)=-3$, I get that $x=2$ on the simplified expression. However, in the original expression $f(x)=-3$ is undefined. So I'm a bit confused by this.
Can I say that $f(2)$ is defined or not?
 A: In simplifying $f(x)$, one would probably take the following steps:
\begin{align}
    f(x) &= \frac{\frac{x-14}{x-2}-1}{7+ \frac{4}{x-2} } \\
    &= \frac{\frac{x-14}{x-2}-1}{7+ \frac{4}{x-2} } \times \frac{x-2}{x-2} \\
    &= \dots
\end{align}
In doing this, an implicit assumption one has already made is that $\frac{x-2}{x-2}$ is well-defined (i.e. $x \neq 2$).
TL;DR: No, $f(2)$ is not defined.
A: The function we have obtained by manipulation
$$g(x)=\frac{-12}{7x-10}$$
is identical to $f(x)$ over all the domain excluding the point $x=2$ for which $g(x)$ is defined but $f(x)$ is not.
Note also that
$$\lim_{x\to 2} f(x) = g(2)=-3$$
and we can use that fact to eliminate the discontinuity for $f(x)$ at the point $x=2$ which is a removable singularity for the function.
Therefore $f(2)$ for the given function is not defined but we can redefine the function as follows
$$f^*(x)=\begin{cases}\frac{\frac{x-14}{x-2}-1}{7+ \frac{4}{x-2} }\quad \text{for} \quad x\neq 2\\-3\quad \quad\text{for}\quad x=2\end{cases}$$
removing the discontinuity at the point $x=2$ and by this definition we have $f^*(x)=g(x)$.
To conclude we have that:

*

*$f(x)=3$ has not solution

*$g(x)=f^*(x)=3$ has one solution at $x=2$
A: $$f(x)=\frac{\frac{x-14}{x-2}-1}{7+ \frac{4}{x-2} }\\=\frac{\frac{-12}{x-2}}{\frac{7x-10}{x-2}}\\=\frac{12}{10-7x}\times\frac{x-2}{x-2}$$
Thus, $f$'s domain is some subset of $$\mathbb R \setminus\left\{\frac{10}7,2\right\}.$$
If $f$'s domain is explicitly specified, then we can get rid of the $\frac{x-2}{x-2}$ and simply write $$f(x)=\frac{12}{10-7x}.$$
A: You can just state that $2$ is not a valid input: $f(x)=\frac{-12}{7x-10}$ for $x \neq2$.
I don't think it's standard notation, but $f(x\neq2)=\frac{-12}{7x-10}$ would probably be understood.
