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Question Number 40106 by maxmathsup by imad last updated on 15/Jul/18
study and give the graph for the function f(x)= (x/(x−1)) e^(1/x)
studyandgivethegraphforthefunctionf(x)=xx−1e1x
Answered by math khazana by abdo last updated on 28/Jul/18
D_f =R−{0,1} =]−∞,0[∪]0,1[∪]1,+∞[ lim_(x→−∞) f(x)=lim_(x→−∞) (x/(x−1))e^(1/x) =1 y=1 is assymtote at −∞ lim_(x→+∞) f(x)=lim_(x→+∞) (x/(x−1)) e^(1/x) =1 so y=1 is assymptote at+∞ lim_(x→1^− ) f(x) =−∞ and lim_(x→1^+ ) f(x)= +∞ lim_(x→0^+ ) f(x)=lim_(x→0^+ ) (x/(x−1)) e^(1/x) chang=(1/x)=t =lim_(t→+∞) (1/(t((1/t)−1))) e^t =lim_(t→+∞) (1/(1−t)) e^t =lim_(t→+∞) (e^t /t) (1/((1/t)−1)) =−∞ lim_(x→0^− ) f(x) =lim_(x→0^− ) (x/(x−1)) e^(1/x) (ch.(1/x)=t) =lim_(t→−∞) (1/(t((1/t)−1))) e^t =lim_(t→−∞) −(e^t /t)=0 we have f^′ (x)=((x/(x−1)))^′ e^(1/x) +((x/(x−1)))(e^(1/x) )^′ = −(1/((x−1)^2 )) e^(1/x) −(1/x^2 )((x/(x−1)))e^(1/x) =−(1/((x−1)^2 )) e^(1/x) − (1/(x(x−1))) e^(1/x) =−(1/((x−1))) e^(1/x) { (1/x) +1} =−((x+1)/x)(1/((x−1)))e^(1/x) =−((x+1)/(x(x−1))) e^(1/x) ....be continued...
Df=R−{0,1}=]−∞,0[∪]0,1[∪]1,+∞[limx→−∞f(x)=limx→−∞xx−1e1x=1y=1isassymtoteat−∞limx→+∞f(x)=limx→+∞xx−1e1x=1soy=1isassymptoteat+∞limx→1−f(x)=−∞andlimx→1+f(x)=+∞limx→0+f(x)=limx→0+xx−1e1xchang=1x=t=limt→+∞1t(1t−1)et=limt→+∞11−tet=limt→+∞ett11t−1=−∞limx→0−f(x)=limx→0−xx−1e1x(ch.1x=t)=limt→−∞1t(1t−1)et=limt→−∞−ett=0wehavef′(x)=(xx−1)′e1x+(xx−1)(e1x)′=−1(x−1)2e1x−1x2(xx−1)e1x=−1(x−1)2e1x−1x(x−1)e1x=−1(x−1)e1x{1x+1}=−x+1x1(x−1)e1x=−x+1x(x−1)e1x….becontinued…

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