I have a graph function:

enter image description here

How to understand the word 'invertible' in this case?

  • $\begingroup$ $f$ is invertible if there is a function $g$ with domain of $f$ being range of $g$ and vice versa; such, that $f(g(x))=x$ for every $x$ in domain of $g$ and $g(f(x))=x$ for every $x$ in domain of $f$. $g$ is usually written down as $f^{-1}$. $f$ is one-to-one on its domain. So it's asked to narrow the domain of $\tan$ to an interval where it's one-to-one. $\endgroup$ Commented Jun 5, 2015 at 5:17
  • $\begingroup$ $\tan$ (or any function) is invertible on those intervals that pass the horizontal line test. $\endgroup$
    – John Joy
    Commented Jun 6, 2015 at 19:22

1 Answer 1


If a function has an inverse, then there is at most one $x$-value for each $y$-value. However, the tangent function is periodic with period $\pi$. Hence, at each value for which $f(x) = \tan x$ is defined, $f(x + n\pi) = \tan x$ for each integer $n$.


Consequently, the function $f(x) = \tan x$ does not have an inverse. Therefore, if we want to obtain an inverse, we must restrict its domain to form a new function in which there is at most one $x$-value for each $y$-value. From examining the graph of the tangent function, we see that in each interval of the form $$\left((2k - 1)\frac{\pi}{2}, (2k + 1)\frac{\pi}{2}\right)$$ where $k$ is an integer, the tangent function assumes every value in its range. Moreover, in each such interval, each $y$-value is achieved exactly once. Hence, we can create an invertible function by restricting the domain tangent function to one such interval. The standard way to do this is to restrict the domain to $-\frac{\pi}{2} < x < \frac{\pi}{2}$, which yields the invertible function $$g(x) = \tan x, -\frac{\pi}{2} < x < \frac{\pi}{2}$$


As you can see from inspecting its graph, the domain and range of $g(x)$ are, respectively, \begin{align*} D_g & = \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\\ R_g & = (-\infty, \infty) \end{align*} and there is exactly one value of $x$ for each value of $y$. Consequently, $g(x)$ has an inverse $g^{-1}(x) = \arctan x$ with domain and range \begin{align*} D_{g^{-1}} & = (-\infty, \infty)\\ R_{g^{-1}} & = \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{align*}



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