Tangents and differentiability

Question

Was just going through my single variable calculus notes recently when I came across this interesting article on the relation between differentiabilty of a function. I missed some of the points my professor made and hence a few doubts remain in my mind -

1. Can a function be differentiable at a point and still have no tangent at that point?

2. Can a function have a tangent at a point and still be non-differentiable at the point?

Answer 1

Differentiability at a point, for a real-valued function of one variable, is the same as the existence of a tangent line at that point, except for one case: If the tangent line is vertical, then the function is not differentiable at that point. It's because a vertical line isn't represented by any linear function, and differentiability is really about looking locally like a linear function.

Thus, the answer to your first question is "no", and the answer to your second question is "yes, consider $y=x^{1/3}$ at $(0,0)$"

Answer 2

If a function is differentiable at some point $(x_0,y_0)$ with slope $m$, then the tangent is $y-y_0=m(x-x_0)$. And so, the first proposition is false.

As for the second proposition, it is true, yes a function can have a tangent without being differentiable. Consider the function $$y=\sqrt{25-x^2}$$

It has tangents $x=5$ and $x=-5$ but it is not differntiable at these points of tangency.