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Question Number 28138 by abdo imad last updated on 21/Jan/18
studie and?give the graph for the function f(x)= e^x −x^e .
studieand?givethegraphforthefunctionf(x)=ex−xe.
Commented by abdo imad last updated on 26/Jan/18
we have f(x)= e^x −e^(elnx) ⇒ D_f =]0,+∝[ lim_(x→0^+ ) f(x)= 1 and lim_(x→+∞) f(x)=lim_(x→+∞) e^x ( 1−e^(−x+elnx) ) =lim_(x→+∞) e^x (1 −e^(−x(1−((elnx)/x))) )=lim _(x→+∞^ ) e^x =+∞ lim_(x→+∞) ((f(x))/x)= lim_(x→+∞) (e^x /x) − x^(e−1) =lim_(x→+∞) (e^x /x)( 1− e^(−x) x^e )=lim_(x→+∞) (e^x /x) =+∞ so the graph of f have a parabolic diction at +∞ f^′ (x)= e^x − e x^(e−1) = e^x (1− e^(−x+1) e^((e−1)lnx) ) = e^x ( 1−e^(−x+1 +(e−1)lnx) ) the sine off^′ (x) is thesine of ψ(x)= 1− e^((e−1)lnx −x+1) ψ^′ (x)= −( ((e−1)/x) −1) e^((...)) =−( ((e−1−x)/x))e^((...)) =((x−(e−1))/x) e^((...)) and ψ^′ (x)=0 ⇔ x=e−1 and f^′ (e−1)=e^(e−1) (1− e^(−e+2) e^((e−1)ln(e−1)) ).....
wehavef(x)=ex−eelnx⇒Df=]0,+∝[limx→0+f(x)=1andlimx→+∞f(x)=limx→+∞ex(1−e−x+elnx)=limx→+∞ex(1−e−x(1−elnxx))=limx→+∞ex=+∞limx→+∞f(x)x=limx→+∞exx−xe−1=limx→+∞exx(1−e−xxe)=limx→+∞exx=+∞sothegraphoffhaveaparabolicdictionat+∞f′(x)=ex−exe−1=ex(1−e−x+1e(e−1)lnx)=ex(1−e−x+1+(e−1)lnx)thesineoff′(x)isthesineofψ(x)=1−e(e−1)lnx−x+1ψ′(x)=−(e−1x−1)e(…)=−(e−1−xx)e(…)=x−(e−1)xe(…)andψ′(x)=0⇔x=e−1andf′(e−1)=ee−1(1−e−e+2e(e−1)ln(e−1))…..

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