scoladan

Verified answer

Bonjour,

1)

a) f(t) = (t-2)/t(t-1)

2 pôles simples 0 et 1

Partie entière = 0 car degré numérateur < degré dénominateur

==> f(t) = A/t + B/(t-1)

tf(t) = A + Bt/(t-1)

==> A = lim tf(t) quand t -->0

==> A = lim t(t-2)/t(t-1) = lim (t-2)/(t-1) = -2/-1 = 2

Et :

(t-1)f(t) = A(t-1)/t + B

==> B = lim (t-1)f(t) quand t-->1

Soit B = lim (t-1)(t-2)/t(t-1) = lim (t-2)/t = -1/1 = -1

==> f(t) = 2/t - 1/(t-1)

b) idem

c) f(t) = t^3/(t^2 - 3t + 2) = t^3/(t-1)(t-2)

2 pôles simples 1 et 2

Division euclidienne :

t^3 = t(t^2 - 3t + 2) + 3t^2 - 2t

= t(t^2 - 3t + 2) + 3(t ^2 - 3t + 2) + 7t - 6

==> f(t) = t + 3 + A/(t-1) + B/(t-2)

A = lim (t-1)f(t) quand t-->1 = lim t^3/(t-2) = -1

B = lim (t-2)f(t) quand t-->2 = lim t^3/(t-1) = 8

==> f(t) = t + 3 - 1/(t-1) + 8/(t-2)


2) f(-x) = - 2x/(x^2 + 1) - Arctan(-x) = -2x/(x^2 +1) + Arctan(x) = - f(x)

==> f impaire ==> Symétrie / O ==> Etude sur [0, +infini[

2) lim f(x) qd x-->+ infini = lim (-Arctan(x)) = - pi/2

3) f'(x) = [2(x^2+1) - 4x^2]/(x^2 + 1)^2 - 1/(x^2 + 1)

= 1/(x^2 + 1)^2 [2x^2 - 4x^2 + 2 - 1 - x^2]

= (1 - 3x^2)/(x^2+1)^2

S'annule sur [0, +infini[ pour x = V(3)/3

4)

5) y = f'(?)(x-?) + f(?) (on voit pas l'abcisse en question)