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Question Number 90940 by  M±th+et+s last updated on 27/Apr/20
f(x)=(x)^(1/3) is there an inflection point when x=0
$${f}\left({x}\right)=\sqrt[{\mathrm{3}}]{{x}}\:\:{is}\:{there}\:{an}\:{inflection}\:{point} \\ $$$${when}\:{x}=\mathrm{0} \\ $$
Answered by MJS last updated on 27/Apr/20
if we stay in R ⇒ ((−x))^(1/3) =−(x)^(1/3) f(x)=(x)^(1/3) =x^(1/3) f′(x)=(1/3)x^(−2/3) is not defined for x=0 lim_(x→0^− ) f′(x) =−∞ but lim_(x→0^+ ) f′(x) =+∞ f′′(x)=−(2/9)x^(−5/3) is not defined for x=0 lim f′′(x) =+∞ but lim f′′(x) =−∞ the curvature is c(x)=((f′′(x))/((f′^2 (x)+1)^(3/2) ))=−((6x^(1/3) )/((9x^(4/3) +1)^(3/2) )) c(x) { ((>0; x<0)),((=0; x=0)),((<0; x>0)) :} ⇒ we have an inflection point we cannot find in the usual way it′s similar to the problem to find the equation y=ax+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
$$\mathrm{if}\:\mathrm{we}\:\mathrm{stay}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{x}}=−\sqrt[{\mathrm{3}}]{{x}} \\ $$$${f}\left({x}\right)=\sqrt[{\mathrm{3}}]{{x}}={x}^{\mathrm{1}/\mathrm{3}} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\mathrm{2}/\mathrm{3}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{x}=\mathrm{0} \\ $$$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:{f}'\left({x}\right)\:=−\infty\:\mathrm{but}\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{f}'\left({x}\right)\:=+\infty \\ $$$${f}''\left({x}\right)=−\frac{\mathrm{2}}{\mathrm{9}}{x}^{−\mathrm{5}/\mathrm{3}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{x}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{lim}\:{f}''\left({x}\right)\:=+\infty\:\mathrm{but}\:\mathrm{lim}\:{f}''\left({x}\right)\:=−\infty \\ $$$$\mathrm{the}\:\mathrm{curvature}\:\mathrm{is}\:{c}\left({x}\right)=\frac{{f}''\left({x}\right)}{\left({f}'^{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }=−\frac{\mathrm{6}{x}^{\mathrm{1}/\mathrm{3}} }{\left(\mathrm{9}{x}^{\mathrm{4}/\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${c}\left({x}\right)\:\begin{cases}{>\mathrm{0};\:{x}<\mathrm{0}}\\{=\mathrm{0};\:{x}=\mathrm{0}}\\{<\mathrm{0};\:{x}>\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{an}\:\mathrm{inflection}\:\mathrm{point}\:\mathrm{we}\:\mathrm{cannot} \\ $$$$\mathrm{find}\:\mathrm{in}\:\mathrm{the}\:\mathrm{usual}\:\mathrm{way} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{similar}\:\mathrm{to}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{equation} \\ $$$${y}={ax}+{b}\:\mathrm{for}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{line}.\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{switch} \\ $$$$\mathrm{to}\:{x}=\mathrm{constant}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}. \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{inflection}\:\mathrm{point}\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{using} \\ $$$${g}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)={x}^{\mathrm{3}} \\ $$$$\mathrm{of}\:\mathrm{course}\:\mathrm{an}\:\mathrm{inflection}\:\mathrm{point}\:\mathrm{is}\:\mathrm{still}\:\mathrm{an} \\ $$$$\mathrm{inflection}\:\mathrm{point}\:\mathrm{after}\:\mathrm{changing}\:{x}\rightleftarrows{y} \\ $$
Commented by  M±th+et+s last updated on 27/Apr/20
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$

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