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Can you have a vertical tangent where a function is undefined? For example, the function $y = 1/x$ is undefined at $x=0$, but that's where the denominator of its derivative is equal to $0$.

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    No. A line can only be tangent to a point on a curve. There is no point on the graph of $f(x) = \frac 1x$ with an x-coordinate of $0$. – MathematicsStudent1122 Jul 13 '17 at 01:03
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    You can, however, have vertical tangents on other functions where the curve does properly pass through $0$. For example, consider $f(x) = x^{\frac 13}$ at $x=0$. Nevertheless, this function is not differentiable at $x=0$ since we do not consider infinite derivatives to be well-defined. – MathematicsStudent1122 Jul 13 '17 at 01:06

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You cannot have a tangent to a curve at any point that's not on the curve, nor a tangent to a graph of a function at a point not in its domain.

The reciprocal function $x\mapsto1/x$ has the set $\mathbb R\smallsetminus \{0\}$ as both its domain and its image.

However, one can do something like this: consider the set $\mathbb R\cup\{\infty\}$ where $\infty$ is neither $+\infty$ nor $-\infty$ but a single $\infty$ that is approached by going in either the positive or the negative direction. Then $x\mapsto1/x$ is a continuous function whose domain and whose image are both $\mathbb R\cup\{\infty\}.$ One can then consider how to define "tangent line" in a way that is consistent with the definitions used when working only with real numbers, and if this is successful, one should find that the $y$-axis is a tangent line to the graph of this function at $x=0$ and the $x$-axis is its tangent line at $x=\infty.$

Michael Hardy
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