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$f:\{x\in\mathbb{R}\colon x\neq\frac{3}{2}\}\longrightarrow\mathbb{R}$ defined by $\displaystyle f(x)=\frac{4x+1}{2x-3}$.

This is for an introductory discrete math class. To tell the truth, I'm not sure how to definitively find this out. I know that the function would normally be undefined at $x=\frac{3}{2}$, and therefore there will be a point in $\mathbb{R}$ not in the range... but to find that point I'm not sure what to do.

Wolfram Alpha says $\{x\in\mathbb{R}:x\neq 2\}$... but how did it come up with 2? The only thing I can think of is to try and find the left hand and right hand limits, and unless I'm thinking wrong, I don't see how 2 comes up.

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Asking what is the range is the same as asking for what values of $t\in\mathbb{R}$ there exists a solution to $$f(x)=\frac{4x+1}{2x-3}=t$$ Multiplying and expressing $x$, we have: $$4x+1=(2x-3)t\Rightarrow 3t+1=(2t-4)x\Rightarrow x=\frac{3t+1}{2t-4}$$ So, what are the possible values of $t$?

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  • $\begingroup$ I'm not following - how did you get from $4x+1=(2x-3)t$ to $3t+1=(2t-4)x$? $\endgroup$ – agent154 Mar 7 '13 at 17:52
  • $\begingroup$ OK, scratch that - I see it now. Never would have thought of that. Thanks $\endgroup$ – agent154 Mar 7 '13 at 17:56
  • $\begingroup$ This 'inversion' also has a geometric interpretation: a function of the form $y=\dfrac{ax+b}{cx+d}$ (with $ad-bc=0$, as @DanShved says in a different answer) represents a hyperbola with its asymptotes parallel to the axes. The fact that inverting it gives a function of the same form (i.e., that $x=f^{-1}(y) = \dfrac{gy+h}{jy+k}$) then says that reflecting this hyperbola about the line $y=x$ (or equivalently, rotating it $45^\circ$) yields another hyperbola with asymptotes parallel to the axes. $\endgroup$ – Steven Stadnicki Mar 7 '13 at 18:40
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Try the following: Let $y = f(x)$ and solve for $x$ (in terms of $y$). This will give you a formula of a form similar to $f$ that will answer the question.

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I know that this is not exactly the answer the OP is looking for, but still seems worth mentioning. Lots of guys will say that the answer is $\mathbb{R} \setminus \{2\}$ right away just after looking at the question. How do they do that?

Rational functions like this: $$ f(x) = \frac{ax + b}{cx + d} $$ have lots of nice properties if $ad - bc \neq 0$ (If $ad - bc=0$, then such a function is either a constant or nowhere-defined). These properties look even nicer if we introduce infinity. Infinity is just a symbol that we add to the set of reals to get $\bar{\mathbb{R}} = \mathbb{R} \cup \{\infty\}$. The rules for dealing with infinity are intuitive here: when $cx + d = 0$ we set $f(x)=\infty$, and we also set $f(\infty)=\frac{a}{c}$ (that is the limit of $f(x)$ as $x$ goes to infinity).

Nice fact: with these rules, $f$ is a bijection from $\bar{\mathbb{R}}$ to itself. It is defined everywhere, its range is the whole $\bar{\mathbb{R}}$, and for each $y$ in $\bar{\mathbb{R}}$ there is exactly one $x$ such that $f(x)=y$. Of course, this needs to be proved, but if you already know this - your problem becomes much easier.

In your problem, $f$ is defined on the whole $\bar{\mathbb{R}}$ except two points: $\frac{3}{2}$ and $\infty$. If we extend $f$ to the whole $\bar{\mathbb{R}}$ according to our rules for dealing with infinity, we will get a bijection $\bar{f}:\ \bar{\mathbb{R}} \to \bar{\mathbb{R}}$ such that $\bar{f}(\frac{3}{2}) = \infty$ and $\bar{f}(\infty) = \frac{4}{2}=2$. So, the two values that $f$ does not have in its range are $\infty$ and $2$, hence the answer.

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Let $$y=\frac{4x+1}{2x-3}$$ and express $x$ in terms of $y$. Now, check for what values of $y$ in co-domain $x$ doesn't lie in domain and your range then would be Co-domain $- $ set of those points.

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