Etudes de fonctions:
(niveau II
es)
Etudier les fonctions f suivantes:
( domf, domf0, limites aux bornes de domf , f0(x) sous forme factorisée):
1. f (x) = x 2 x + 2 2. f (x) = x − 1 +x + 19 3. f (x) = 4x 2+ 4x + 5 4x + 2 4. f (x) = x 2+ 2x + 2 x − 1 5. f (x) = −x 2 − 2x − 1 x + 3 6. f (x) = x 2 |x| + 2 7. f (x) = −2x + 1 + 2 x − 1 8. f (x) = |x + 1| + 1 x − 1 9. f (x) =(x − 2) 2 x2+ 2 10. f (x) = x 2− 4x + 6 (x − 2)2 11. f (x) = 2x 2− 3 x2− 1
Solutions:
1. f (x) = x 2 x + 2 domf = R\{−2} =domf0 lim x→−∞ µ x2 x + 2 ¶ = −∞ lim x→+∞ µ x2 x + 2 ¶ = +∞ lim x→−2− µ x2 x + 2 ¶ = −∞ lim x→−2+ µ x2 x + 2 ¶ = +∞ f0(x) = µ x2 x + 2 ¶0 = x(x + 4) (x + 2)2 2. f (x) = x − 1 + 9 x + 1 domf = R\{−1} =domf0 lim x→−∞ µ x − 1 + 9 x + 1 ¶ = −∞ lim x→+∞ µ x − 1 + 9 x + 1 ¶ = +∞ lim x→−1− µ x − 1 + x + 19 ¶ = −∞ lim x→−1+ µ x − 1 +x + 19 ¶ = +∞ f0(x) = µ x − 1 + x + 19 ¶0 = (x − 2) (x + 4) (x + 1)2 3. f (x) = 4x 2+ 4x + 5 4x + 2 domf = R\{−1} = domf0lim x→−∞ µ 4x2+ 4x + 5 4x + 2 ¶ = −∞ lim x→+∞ µ 4x2+ 4x + 5 4x + 2 ¶ = +∞ lim x→−1 2 − µ 4x2+ 4x + 5 4x + 2 ¶ == −∞ lim x→−1 2 + µ 4x2+ 4x + 5 4x + 2 ¶ = +∞ f0(x) = µ 4x2+ 4x + 5 4x + 2 ¶0 =(2x − 1) (2x + 3) (2x + 1)2 4. f (x) = x 2+ 2x + 2 x − 1 domf = R\{1} = domf0 lim x→−∞ µ x2+ 2x + 2 x − 1 ¶ = −∞ lim x→+∞ µ x2+ 2x + 2 x − 1 ¶ = +∞ lim x→1− µ x2+ 2x + 2 x − 1 ¶ = −∞ lim x→1+ µ x2+ 2x + 2 x − 1 ¶ = +∞ f0(x) = µ x2+ 2x + 2 x − 1 ¶0 = x 2− 4 − 2x (x − 1)2 5. f (x) = −x 2− 2x − 1 x + 3 domf = R\{−3} = domf0 lim x→−∞ µ −x2− 2x − 1 x + 3 ¶ = +∞ lim x→+∞ µ −x2− 2x − 1 x + 3 ¶ = −∞ lim x→−3− µ −x2− 2x − 1 x + 3 ¶ = +∞
lim x→−3+ µ −x2− 2x − 1 x + 3 ¶ = −∞ f0(x) = µ −x2− 2x − 1 x + 3 ¶0 = −(x + 1) (x + 5) (x + 3)2 6. f (x) = x 2 |x| + 2 domf = R domf0= R lim x→−∞ µ x2 |x| + 2 ¶ = +∞ = lim x→+∞ µ x2 |x| + 2 ¶ f (x) = x2 −x + 2 si x ≤ 0 x2 x + 2 si x ≥ 0 f est d´erivable en0 et f0(0) = 0!! f0(x) = −x(x − 4) (x − 2)2 si x ≤ 0 x(x + 4) (x + 2)2 si x ≥ 0 7. f (x) = −2x + 1 + 2 x − 1 dom f = R\{1} = dom f0 lim x→−∞ µ −2x + 1 + 2 x − 1 ¶ = +∞ lim x→1− µ −2x + 1 + 2 x − 1 ¶ = −∞ lim x→1+ µ −2x + 1 + 2 x − 1 ¶ = +∞ lim x→+∞ µ −2x + 1 + 2 x − 1 ¶ = −∞ f0(x) = µ −2x + 1 + 2 x − 1 ¶0 = −2(x 2− 2x + 2) (x − 1)2
8. f (x) = µ |x + 1| + 1 x − 1 ¶ dom f = R\{1} dom f0= R\{1, −1} lim x→−∞ µ |x + 1| + 1 x − 1 ¶ = +∞ = limx →+∞ µ |x + 1| + 1 x − 1 ¶ lim x→1− µ |x + 1| + 1 x − 1 ¶ = −∞ lim x→1+ µ |x + 1| + 1 x − 1 ¶ = +∞ f (x) = −x − 1 + 1 x − 1 si x ≤ −1 x + 1 + 1 x − 1 si x ≥ −1 et x 6= 1 f0(x) = −1 − 1 (x − 1)2 = − x2− 2x + 2 (x − 1)2 si x < −1 1 − 1 (x − 1)2 = (x − 2) x (x − 1)2 si x > −1 et x 6= 1 9. f (x) =(x − 2) 2 x2+ 2 dom f = R =domf0 lim x→−∞ Ã (x − 2)2 x2+ 2 ! = 1 = lim x→+∞ Ã (x − 2)2 x2+ 2 ! f0(x) = Ã (x − 2)2 x2+ 2 !0 = 4 (x − 2) (x + 1) (x2+ 2)2 10. f (x) = x 2− 4x + 6 (x − 2)2 dom f = R\{2} = domf0 lim x→−∞ Ã x2− 4x + 6 (x − 2)2 ! = 1 = lim x→+∞ Ã x2− 4x + 6 (x − 2)2 ! lim x→2− Ã x2− 4x + 6 (x − 2)2 ! = +∞ = lim x→2+ Ã x2− 4x + 6 (x − 2)2 !
f0(x) = Ã x2− 4x + 6 (x − 2)2 !0 = − 4 (x − 2)3 11. f (x) = 2x 2− 3 x2− 1 dom f = R\{−1; 1} = domf0 lim x→−∞ µ 2x2− 3 x2− 1 ¶ = 2 = lim x→+∞ µ 2x2− 3 x2− 1 ¶ lim x→−1− µ 2x2− 3 x2− 1 ¶ == −∞ lim x→−1+ µ 2x2− 3 x2− 1 ¶ = +∞ lim x→1− µ2x2− 3 x2− 1 ¶ = +∞ lim x→1+ µ2x2− 3 x2− 1 ¶ = −∞ f0(x) = µ2x2− 3 x2− 1 ¶0 = 2x (x2− 1)2
Mise en page des énoncés et
rédaction et mise en page des solutions: Christophe THEIS (Iere B 4, LCD 2004)
