TITLE: How can a "proper" function have a vertical slope?
QUESTION [44 upvotes]: Plotting the function $f(x)=x^{1/3}$ defined for any real number $x$ gives us:

Since $f$ is a function, for any given $x$ value it maps to a single y value (and not more than one $y$ value, because that would mean it's not a function as it fails the vertical line test).
This function also has a vertical tangent at $x=0$. 
My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $x$. How is it possible for a "proper" function to have a vertical tangent?
As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function?

REPLY [2 votes]: The derivative of the function is
$$ y'=\frac{1}{3x^{2/3}} $$
$y'(0)$ is undefined due to division by zero, but
$$ lim_{x \to 0} \,\, y' = \pm\infty $$
Alternatively, $x$ can be expressed as a function of $y$.
$$
\begin{aligned}
x &= y^3 \\
x' &= 3y^2 \\
x'(0) &= 0 \\
\end{aligned}
$$
Since $x'(0) = 0$, the original function ($ y = x^{1/3} $) must have a vertical tangent at $x=0$, because $x$ does not change when $y$ changes around that point.