L is tangent to y = f(x) at P(x_P, f(x_P)) <=> P is on L and L has slope f'(x_P)

[richardt]

L is tangent to y = f(x) at P(x_P, f(x_P)) <=> P is on L and L has slope f'(x_P)

Greetings: Supposedly, line L is tangent to function y = f(x) at P(x_P, f(x_P)) if and only if P is on L and L has slope f'(x_P). But for f(x) = x^1/3, f'(0) fails to exist. Hence f has no tangent at (0, 0). But this contradicts the assertion that f has a vertical tangent at the origin. Which is it?

Thank you.

R. Briggaman

 

[ksdhart]

I believe it would be more accurate to say that the slope of a tangent line to a curve is equal to the limit of the derivative function as x approaches whatever value. For most cases, this would produce the same behavior as you'd expect, but it handles odd cases like this as well. So, here the derivative is:

\(\displaystyle \frac{d}{dx}\left(x^{\frac{1}{3}}\right)=\frac{1}{3x^{\frac{2}{3}}}\)

The limit of f'(x) as x approaches 0 is:

\(\displaystyle \lim _{x\to 0}\left(\frac{1}{3x^{\frac{2}{3}}}\right)=\infty\)

We can interpret this as meaning that the slope of the tangent line at x=0 is infinity, which means it's a vertical line. And that fits with an observation of the graph, which shows a vertical tangent line at x=0.

 

[richardt]

Thank you kindly. That works for me.