# does $x^{ \frac{1}{3} }$ have a tangent line at x=0? [duplicate]

it is not differentiable but it should have a tangent line, x=0 as in the y-axis, still, no?

how is the notion of the derivative and the existence of a tangent line reconciled here?

(yes, I know there are repeated versions of this question but I haven't seen one talk about the tangent line perspective)

• Are you asking if there can be a tangent at a point of inflection (depends on definition of tangent), or if a vertical line can be a tangent (yes, but you may need to describe it in a different way)? – Henry Oct 19 '19 at 0:21
• The derivative does not exist but it has a tangent line. If the derivative of a function exists then its value represents the slope of a line tangent to the graph of the function. That does not mean that if the derivative doesn't exist, neither does the tangent line. – John Douma Oct 19 '19 at 0:22
• @Henry I guess both questions you posed are something I am curious with the answers to. It is just that whenever calculus derivatives are talked about, the concept linked to it is the existence of a tangent line. Is there a theorem about the existence of a tangent line maybe (this might go into analysis, which is okay by me) – user29418 Oct 19 '19 at 0:26
• Vertical lines do not have a slope and in this case the tangent line is vertical. – Mohammad Riazi-Kermani Oct 19 '19 at 0:28
• @JohnDouma yes if the derivative exists, the tangent line exists. but when the derivative does not exist, the tangent line's existence is inconclusive, is there a way to check the existence of a tangent line (including a vertical one) not by intuition or picture but analytically/algebra or rigorously? – user29418 Oct 19 '19 at 0:29