The graph of a function cannot have a vertical tangent line at a point of inflection. Is this false?

Answer 1

Consider y = the cube root of x.

Think about where the inflection point is, and what the first derivitive is at that point.

Answer 2

It has been a while since I have thought of this stuff, but I'll try.

A point of inflection is one where f''(x) = 0.
But if the tangent is vertical then f'(x) is undefined (infinite), so its derivative cannot be 0. Make sense?

No, DanE, while I am not 100% sure of my answer, your example is not right. A parabola opening to the side is not a function because it is dual valued at most points. If you just take the upper branch (or lower branch) it is a function, but it is not an inflection point at the vertex.

But I think your idea is right (and mine is wrong): instead of a parabola take a cubic on its side: y = x^(1/3). This is a function, f'(0) is vertical, and it is an inflection point. My mistake is that it is not necessary for a point of inflection to have 2nd derivative of 0 at p, it needs to have f''(p + e) and f''(p - e) to have different signs, which can also happen if f''(p) is undefined at the point.

Answer 3

A graph of a function CAN have a vertical tangent line at a point of inflection.
Think of a graph of a parabola that opens to the left. The tangent to the inflection point will go up and down.

Answer 4

Do we define an inflection point as a point where the curvature changes sign? If so, the tangent at that point can be vertical (as other responders have noted, using the example y = x^(1/3)
(or x = y^3).

Or do we define an inflection point as a point where the second derivative of the function (of x) is 0? In that case, I don't think the tangent can be vertical at such a point. It can be horizontal, of course; that's just the inverse of the situation described in the preceding paragaph.