Does $f(x)=x \ln (x)$ have a vertical tangent

Question

Does $f(x)=x \ln (x)$ have a vertical tangent

We have domain of $f(x)$ as $(0 ,\: \: \infty)$

Now

$$f'(x)=1+\ln x$$

we have

$$\lim_{x \to 0^+} f'(x) =-\infty$$

Can we say $f$ has vertical tangent at origin?

Answer 1

No, for there is no point there on the graph for which you could build such a tangent line ($x=0$ is outside the domain, as you've noted). The function $f(x) =x^{2/3}$ exhibits the behavior you are asking about.

Answer 2

The function is not defined at $0$, so it cannot have a tangent there. However, if you extend it by declaring $$ f(x)=\begin{cases} x\ln\lvert x\lvert & x\ne0 \\[4px] 0 & x=0 \end{cases} $$ then the function is continuous at $0$ and differentiable for $x\ne0$; since $$ \lim_{x\to0}f'(x)=\lim_{x\to0}(\ln|x|+1)=-\infty $$ you can indeed conclude that the function has vertical tangent at $0$.

_{It's not the only way to extend the function; the important thing is that the value at $x=0$ is $0$, in order to make it continuous. Choosing an odd function doesn't seem a bad idea.}