vertical inflection: does it necessarily follow that 1/ƒ’(xP) = 0?

[richardt]

vertical inflection: does it necessarily follow that 1/ƒ’(xP) = 0?

Greetings: I understand that for a single variable function ƒ_x, point P on ƒ is a vertical inflection point if limit(ƒ’(x)) = ∞ or -∞ as x ⟶ x_P. That said, does it necessarily follow that 1/ƒ’(x_P) = 0? Example: Given ƒ(x) = x^1/3, ƒ has a vertical inflection at (0, 0). Moreover, 1/ƒ’(0) = 3(0)^2/3 = 0. Is this always the case?

Thank you.

Rich B.

 

[ksdhart]

Well, being that the derivative is undefined at the given point, the reciprocal of the derivative is also undefined. If, however, you again look at the limit, what you've said is true. You may remember seeing a rule something like this from your calculus book:

\(\displaystyle \displaystyle\lim _{x\to c}\left(\frac{n}{f\left(x\right)}\right)=0\) if \(\displaystyle \displaystyle \lim _{x\to c}\left(f\left(x\right)\right)=\pm \infty \)

Basically, if you're taking the limit of a fraction where the numerator is any constant n, and the denominator approaches infinity as x approaches c, then the limit is 0.