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let-f-x-x-1-1-x-2-n-study-tbe-derivability-of-f-at-points-0-and-1-n-natural-integr-

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Question Number 40102 by maxmathsup by imad last updated on 15/Jul/18
let f(x) = ((∣x∣)/((1+∣1−x^2 ∣)^n )) study tbe derivability of f at points 0 and 1 (n natural integr)
letf(x)=∣x∣(1+∣1−x2∣)nstudytbederivabilityoffatpoints0and1(nnaturalintegr)
Answered by math khazana by abdo last updated on 26/Jul/18
1) derivability at 0 lim_(h→0^+ ) ((f(0+h) −f(0))/h) =lim_(h→0^+ ) (h/((1+(1−h^2 )^n )) =lim_(h→0^+ ) (h/(h(2+h^2 )^n )) =lim_(h→0^+ ) (1/((2+h^2 )^n )) =(1/2^n ) lim_(h→0^− ) ((f(0+h) −f(0))/h) =lim_(h→0^− ) ((−1)/((2+h^2 )^n )) =((−1)/2^n ) we conclude that f is derivable at left and right at but not derivable at this point.
1)derivabilityat0limh→0+f(0+h)−f(0)h=limh→0+h(1+(1−h2)n=limh→0+hh(2+h2)n=limh→0+1(2+h2)n=12nlimh→0−f(0+h)−f(0)h=limh→0−−1(2+h2)n=−12nweconcludethatfisderivableatleftandrightatbutnotderivableatthispoint.
Answered by math khazana by abdo last updated on 28/Jul/18
2) derivability at 1 lim_(h→0^+ ) ((f(1+h) −f(1))/h) =lim_(h→0^+ ) ((∣1+h∣)/((1+∣1−(1+h)^2 ∣)^n )) −1 =lim_(h→0^+ ) ((1+h)/((1+((1+h)^2 −1))^n )) −1 =lim_(h→0^+ ) ((1+h)/((1+h^2 +2h)^n )) −1 =lim_(h→0^− ) ((1+h−(h+1)^(2n) )/((h+1)^(2n) )) =0 f is derivable at right of 1 lim_(h→0^− ) ((f(1+h)−f(1))/h)=lim_(h→0^− ) ((1+h)/({1+∣1−(1+h)^2 }^n )) −1 =lim_(h→0^− ) ((1+h)/({1+(1−(1+h)^2 }^n )) −1 =lim_(h→0^− ) ((1+h)/({1+(1−h^2 −2h−1)}^n )) −1 =lim_(h→0^− ) ((1+h)/({1−h^2 −2h)^n ))−1 =0 f is derivable at left of 1 we have f_g ^′ (1)=f_d ^′ (1) ⇒ f is derivable at x_0 =1
2)derivabilityat1limh→0+f(1+h)−f(1)h=limh→0+∣1+h∣(1+∣1−(1+h)2∣)n−1=limh→0+1+h(1+((1+h)2−1))n−1=limh→0+1+h(1+h2+2h)n−1=limh→0−1+h−(h+1)2n(h+1)2n=0fisderivableatrightof1limh→0−f(1+h)−f(1)h=limh→0−1+h{1+∣1−(1+h)2}n−1=limh→0−1+h{1+(1−(1+h)2}n−1=limh→0−1+h{1+(1−h2−2h−1)}n−1=limh→0−1+h{1−h2−2h)n−1=0fisderivableatleftof1wehavefg′(1)=fd′(1)⇒fisderivableatx0=1

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