Etude de fonctions:
solutions succintes
1. f (x) = x 2+ x + 2 x2− 3x + 2 domf=R\ {1, 2}=domf’=domf” A.H. : y = 1; A.V. : x = 1; A.V. : x = 2 f0(x) = −4(x2− 2) (x2− 3x + 2)2; f 00(x) = 8(x3− 6x + 6) (x2− 3x + 2)3 -25 -20 -15 -10 -5 0 5 10 -4 -2 2 4x 6 8 2. f (x) = 3 −x12 domf=R∗=domf’=domf” A.H. : y = 3; A.V. : x = 0 f0(x) = 2 x3; f00(x) = − 6 x4 -4 -2 0 2 4 -4 -2 2 x 43. f (x) = 6x x2+ 1 domf=R=domf’=domf” A.H. : y = 0 f0(x) = −6(x2− 1) (x2+ 1)2 ; f 00(x) = 12x(x2− 3) (x2+ 1)3 -3 -2 -1 0 1 2 3 -4 -2 2 x 4 4. f (x) = x 2+ 1 x2− 1 domf=R\ {1, −1}=domf’=domf” A.H. : y = 1; A.V. : x = −1; A.V. : x = 1 f0(x) = −4x (x2− 1)2; f 00(x) = 4(3x2+ 1) (x2− 1)3 -4 -2 0 2 4 -4 -2 2 x 4
5. f (x) = x 2 − 1 x2+ 1 domf=R=domf’=domf” A.H. : y = 1 f0(x) = 4x (x2+ 1)2; f 00(x) = −4(3x2− 1) (x2+ 1)3 -1 -0.5 0 0.5 1 -4 -2 2 x 4 6. f (x) = 1 x2− 2x − 3 domf=R\ {−1, 3}=domf’=domf” A.H. : y = 0; A.V. : x = −1; A.V. : x = 3 f0(x) = −2(x − 1) (x2− 2x − 3)2; f 00(x) = 2(3x2− 6x + 7) (x2− 2x − 3)3 -4 -2 0 2 4 -4 -2 2 x 4 7. f (x) = x 3+ 2x − 3 x2+ 3x − 4 domf=R\ {1, −4}=domf’=domf” cx + d
On trouve:
f (x) = x − 3 +x15x − 152+ 3x − 4
A.V. : x = −1; A.V. : x = 1; A.O. : y = x − 3 f0(x) = x2+ 8x + 1 (x + 4)2 ; f 00(x) = 30 (x + 4)3 -30 -20 -10 0 10 -10 -5 5x 10 8. f (x) = x 3− x + 1 x2 domf=R∗=domf’=domf”
Trouver a,b,c et d réels tels que ∀x ∈domf: f(x) = ax + b +cx + dx2
On trouve: f (x) = x +1 − x x2 A.V. : x = 0; A.O. : y = x f0(x) = x 3 + x − 2 x3 ; f00(x) = − 2(x − 3) x4 -4 -2 0 2 4 -4 -2 2 x 4
9. f (x) = x +1 2+
9x x2− 9
domf=R\ {−3, 3}=domf’=domf”
A.V. : x = −3; A.V. : x = 3; A.O. : y = x +1 2 f0(x) = x2(x2− 27) (x2− 9)2 ; f 00(x) = 18x(x2+ 27) (x2− 9)3 -20 -10 0 10 20 -10 -5 5x 10 10. f (x) = 2 x(x + 1)(x + 2) domf=R∗\ {−1, −2}=domf’=domf”
A.H. : y = 0; A.V. : x = −2; A.V. : x = −1; A.V. : x = 0 f0(x) = −2(3x 2+ 6x + 2) x2(x + 1)2(x + 2)2; f 00(x) = 4(6x4+ 24x3+ 33x2+ 18x + 4) x3(x + 1)3(x + 2)3 -10 -5 0 5 10 -4 -2 2 x 4
