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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(x)=x3isthereaninflectionpointwhenx=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

ifwestayinR⇒−x3=−x3f(x)=x3=x1/3f′(x)=13x−2/3isnotdefinedforx=0limx→0−f′(x)=−∞butlimx→0+f′(x)=+∞f″(x)=−29x−5/3isnotdefinedforx=0limf″(x)=+∞butlimf″(x)=−∞thecurvatureisc(x)=f″(x)(f′2(x)+1)3/2=−6x1/3(9x4/3+1)3/2c(x){>0;x<0=0;x=0<0;x>0⇒wehaveaninflectionpointwecannotfindintheusualwayit′ssimilartotheproblemtofindtheequationy=ax+bforaverticalline.wehavetoswitchtox=constantinthiscase.theinflectionpointcanbefoundusingg(x)=f−1(x)=x3ofcourseaninflectionpointisstillaninflectionpointafterchangingx⇄y

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

god bless you sir

godblessyousir

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