What is a rigorous method/mechanism to find the range of a function $f(x)$? Is it acceptable to find $f^{-1}(x)$ and then its domain? I understand that $f^{-1}(x)$ does not always exist (at least not without limiting the domain of $f(x)$, as is the case of the trigonometric functions) which I imagine can potentially lead to problems when trying to find the range of $f(x)$ in this way.

The other approach I see would be checking the end behaviour of $f(x)$ and potentially considering its local minima and maxima, but I am not convinced that this will aptly handle discontinuities, etc.

Thanks in advance for your help.

  • 2
    $\begingroup$ The "range of $ \ f(x) \ $ is domain of $ \ f^{-1}(x) \ $ " trick only works for one-to-one functions. There isn't really any universal method of finding ranges of functions, since so many possibilities exist. Domains are usually rather easier to characterize. $\endgroup$ Apr 5, 2014 at 0:15

2 Answers 2


After talking to a friend, I was able to work out an answer to my question.

In general, given a function $f$ whose domain is $D\left(f\right)$, its range is $R\left(f\right) = \left\{f\left(x\right) : x \in D\left(f\right) \right\}$.

Suppose that $f$ is a composition of some two functions $g$ and $h$, such that $f\left(x\right) = g\left(h\left(x\right)\right)$. Then we can consider the range of $f$ to be

$R\left(f\right) = \left\{g\left(y\right) : y \in R\left(h\right) \cap D\left(g\right)\right\}$.

Here, we use $g$ to map $R\left(h\right) \stackrel g\to R\left(f\right)$. Thus, so long as we know $R\left(h\right)$ (either a priori or by finding it via this method) and $D\left(g\right)$, we can find $R\left(f\right)$.

To give an example:

Suppose that $f\left(x\right) = \frac{1}{x^2 + 1} - 1$. Then

$R\left(f\right) = \left\{f\left(x\right) : x \in D\left(f\right)\right\} = \left\{y-1 : y \in R\left(\frac{1}{x^2 + 1}\right)\right\}$.


$R\left(\frac{1}{x^2 + 1}\right) = \left\{\frac{1}{y} : y \in R\left(x^2+1\right)\setminus \left\{0\right\}\right\}$

because $D\left(\frac{1}{x}\right) = \mathbb R \setminus \left\{0\right\}$. Again,

$R\left(x^2+1\right) = \left\{y + 1 : y \in R\left(x^2\right)\right\}$

and finally I assume that we trivially recognise that $R\left(x^2\right) = \left[0, \infty\right)$.

It follows that $R\left(x^2+1\right) = \left[1, \infty\right)$ by "adding" $1$ to both endpoints of the interval representing $R\left(x^2\right)$.*

Next, $R\left(\frac{1}{x^2 + 1}\right) = \left(0, 1\right]$ after "taking the reciprocal" of the endpoints. In this case, it turns out that our concern about the domain $D\left(\frac{1}{x}\right)$ was unnecessary because $0 \notin R\left(x^2+1\right)$.

Finally, our original problem $R\left(f\right) = R\left(\frac{1}{x^2 + 1} - 1\right) = \left(-1, 0\right]$ by "subtracting" $1$ from both endpoints.

Sure enough, see its graph below (also see WolframAlpha's analysis of this function).

graph of 1/(x^2 + 1) - 1

*More formally, when you apply a function $f$ to a noninclusive interval, e.g. $\left[a,b\right)$, you are in fact taking the limit $\lim_{x\to b} f\left(x\right)$. That is, $\left\{f\left(x\right) : x \in \left[a,b\right)\right\} = \left[f\left(a\right), \lim_{x\to b^-} f\left(x\right)\right)$. In the cases above, $\lim_{x\to \infty} \left(x+1\right) = \infty$, $\lim_{x\to \infty} \frac{1}{x} = 0$, and $\lim_{x\to 0^+} \left(x-1\right) = -1$.


If you are talking about the real case, then finding the domain of a function is relatively easy. Find the values of $x$ where piece-wise continuous $f(x)$ is nonexistent. Exclude these intervals from $(-\infty,\infty)$ to get the domain. If the domaine is the interval $[a,b]$, then the domain is $[\min f(a),\max f(b)]$.

As an example, consider $f(x)=\sqrt{x}$. This is nonexistent, in the real case, for $x<0$. Thus, the range is simply $[0,\infty)$. The domain, however, is $(-\infty,\infty)$

For complex functions, see this M.SE question.

  • $\begingroup$ Aren't you confounding domain and range ? The range of a function can be difficult to rigorously find. What about $f(x) := \frac{1+\cos x}{1+\sin x}$ ? $\endgroup$
    – Amateur
    Apr 4, 2014 at 23:24
  • $\begingroup$ @Amateur The $f(x)$ you've defined is nonexistent for $\sin x=-1$, i.e., $x=-\pi/2+2n\pi$. $\endgroup$
    – user122283
    Apr 4, 2014 at 23:36
  • $\begingroup$ Yes but that's for the domain. The range of that $f$ is $[0, \infty)$. The question is about the range, not the domain. $\endgroup$
    – Amateur
    Apr 4, 2014 at 23:37
  • $\begingroup$ @Amateur See my edited post. $\endgroup$
    – user122283
    Apr 4, 2014 at 23:39
  • $\begingroup$ Take $f(x) = x^2$ defined on $[-2,2]$. Then you are saying that the range is $[4,4] = \{4\}$ ?! $\endgroup$
    – Amateur
    Apr 4, 2014 at 23:41

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