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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?

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Håkan Mjölk
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    $x=2$ is a removable singularity. It is just a gap for the first expression, but it has been removed in the second one. So, if you use the last expression, I think you can say $f(2)$ is defined. – FeedbackLooper Sep 14 '21 at 14:35
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    Your terminology is a bit confusing. When you say “calculate $f(x)=-3$”, do you mean “solve the equation $f(x)=-3$ (for $x$)”? And when you say “$f(x)=-3$ is undefined”, do you mean “the equation $f(x)=-3$ has no solutions”? – Hans Lundmark Sep 14 '21 at 14:37
  • Sorry, my bad, when I say calculate I mean "solve the equation f(x)=-3 (for x)". And when I say that f(x)=-3 is undefined I mean "the equation f(x)=-3 has no solutions". So you are correct. – Håkan Mjölk Sep 14 '21 at 14:41
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    Anyway, the best person to ask in this case is obviously your teacher! I would suspect that this question was asked in a context where you're supposed to be picky about the domains (maybe a first math course at university?), and in that case $x = 2$ is disqualified (i.e., you don't remove the removable singularity), but that's just a guess. – Hans Lundmark Sep 14 '21 at 14:41
  • So f(x)=-3 is not a solution? – Håkan Mjölk Sep 14 '21 at 14:54
  • @HåkanMjölk. That's correct, $f(x)=-3$ has no solution. – md2perpe Sep 14 '21 at 15:03

4 Answers4

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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.

Tham
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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 $x=2$
user
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$$f(x)=\frac{\frac{x-14}{x-2}-1}{7+ \frac{4}{x-2} }\\=\frac{\frac{-12}{x-2}}{\frac{7x-10}{x-2}}\\=\frac{12}{10-7x}\times\frac{x-2}{x-2}$$

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 $$f(x)=\frac{12}{10-7x}.$$

ryang
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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.

Acccumulation
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