exo15.mw

> Exercice 15

> Question a)

> f:=x->x^2*exp(-2*x);

f := proc (x) options operator, arrow; x^2*exp(-2*x) end proc

D'oł la courbe est au dessus de son asymptote y=x+1/6

> plot([f(x),0],x=-0.8..3,color=[red,blue],thickness=[2,1],discont=true);

[Plot]

> Question b)

> f:=x->(x^3-1)^2;

f := proc (x) options operator, arrow; (x^3-1)^2 end proc

> plot([f(x),0],x=-5..5,color=[red,blue],thickness=[2,1],discont=true);

[Plot]

> Question c)

> f:=x->sqrt(x^2+3*x+1);

f := proc (x) options operator, arrow; sqrt(x^2+3*x+1) end proc

> f(x)=series(f(x),x=infinity,2);f(x)=series(f(x),x=-infinity,2);

(x^2+3*x+1)^(1/2) = x+3/2-5/8/x+O(1/x^2)

(x^2+3*x+1)^(1/2) = -x-3/2+5/8/x+O(1/x^2)

f est en dessous de son asymptote y=x+3/2 en +infini et en dessous de y=-x-3/2 en -infini

> plot([f(x),x+3/2,-x-3/2],x=-8..8,0..10,color=[red,blue,green],thickness=[2,1,1],discont=true);

[Plot]

> Question d)

> f:=x->x^2*ln(abs((x+1)/x));

f := proc (x) options operator, arrow; x^2*ln(abs((x+1)/x)) end proc

> f(x)=series(f(x),x=infinity,4);f(x)=series(f(x),x=-infinity,4);

x^2*ln(abs((x+1)/x)) = x-1/2+1/3/x+O(1/x^2)

x^2*ln(abs((x+1)/x)) = x-1/2+1/3/x+O(1/x^2)

> f est au dessus de son asymptote y=x-1/2 en +infini et en dessous de celle-ci en -infini

> Limit(f(x),x=-1)=limit(f(x),x=-1);

Limit(x^2*ln(abs((x+1)/x)), x = -1) = -infinity

D'oł une  asymptote verticale x=-1

> plot([f(x),x-1/2],x=-5..5,-5..5,color=[red,blue],thickness=[2,1],discont=true);

[Plot]

> Question e)

> f:=x->x^3/(1-2*x^2);

f := proc (x) options operator, arrow; x^3/(1-2*x^2) end proc

> f(x)=series(f(x),x=infinity,3);f(x)=series(f(x),x=-infinity,3);

x^3/(1-2*x^2) = -1/2*x-1/4/x+O(1/x^3)

x^3/(1-2*x^2) = -1/2*x-1/4/x+O(-1/x^3)

> f est en dessous de son asymptote y=-x/2 en +infini et au  dessus de celle-ci en -infini

Il y a de pkus deux asymptotes verticales x=1/racine(2) et  x=-1/racine(2)

> plot([f(x),-x/2],x=1..5,color=[red,blue],thickness=[2,1],discont=true);

[Plot]

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