f-x-x-1-3-is-there-an-inflection-point-when-x-0-

Question Number 90940 by M±th+et+s last updated on 27/Apr/20

f (x) = \sqrt[3]{x} i s t h e r e a n i n f l e c t i o n p o i n t

w h e n x = 0

Answered by MJS last updated on 27/Apr/20

if we stay in R \Rightarrow \sqrt[3]{- x} = - \sqrt[3]{x}

f (x) = \sqrt[3]{x} = x^{1 / 3}

f^{'} (x) = \frac{1}{3} x^{- 2 / 3} is not defined for x = 0

\lim_{x \to 0^{-}} f^{'} (x) = - \infty but \lim_{x \to 0^{+}} f^{'} (x) = + \infty

f^{″} (x) = - \frac{2}{9} x^{- 5 / 3} is not defined for x = 0

\lim f^{″} (x) = + \infty but \lim f^{″} (x) = - \infty

the curvature is c (x) = \frac{f^{″} (x)}{{(f^{' 2} (x) + 1)}^{3 / 2}} = - \frac{6 x^{1 / 3}}{{(9 x^{4 / 3} + 1)}^{3 / 2}}

c (x) {\begin{cases} > 0; x < 0 \\ = 0; x = 0 \\ < 0; x > 0 \end{cases}

\Rightarrow we have an inflection point we cannot

find in the usual way

{it}^{'} s similar to the problem to find the equation

y = a x + b for a vertical line . we have to switch

to x = constant in this case .

the inflection point can be found using

g (x) = f^{- 1} (x) = x^{3}

of course an inflection point is still an

inflection point after changing x ⇄ y

Commented by M±th+et+s last updated on 27/Apr/20

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