Odpowiedź:
zad 1
Wykres w załączniku
f(x) = - 1/(x + 3)
f(x) = - 1 /( - 2 + 3) = - 1/1 = - 1
A - punkt przecięcia z prostą x = - 2 ; ( - 2 , - 1 )
f(x) < - 1
- 1/(x +3) < - 1
1/(x +3) > 1 | * (x + 3)²
x + 3 > x² + 6x + 9
- x² - 6x - 9 + x +3 > 0
- x² - 5x - 6 > 0
a = - 1 , b = - 5 , c = - 6
Δ = b² - 4ac = (- 5)² - 4 * (- 1) * (- 6) = 25 - 24 = 1
√Δ =√1 = 1
x₁ = ( - b - √Δ)/2a =(5 - 1)/(- 2) = - 4/2 = - 2
x₂ = ( - b + √Δ)/2a = (5 + 1)/(- 2) = - 6/2 = - 3
f(x) < - 1 ⇔ x ∈ ( - 3 , - 2 )
zad 2
f(x) = (x⁴ + x³ - 5x² - 5x)/(x³ -x)
założenie:
x³ - x ≠ 0
x(x² -1) ≠ 0
x (x - 1)(x + 1) ≠ 0
x ≠ 0 ∧ x ≠ 1 ∧ x ≠ - 1
D: x ∈ R \ { - 1 , 0 , 1 }
f(x) = (x⁴ + x³ - 5x² - 5x)/(x³- x) = [(x³(x + 1) - 5x(x + 1)]/[x(x + 1)(x - 1)] =
= [(x +1)(x³ - 5x)]/[x(x + 1)(x - 1)] = (x³ - 5x)/[x(x - 1)] = x(x² - 5)/x(x- 1) =
= (x² - 5)/(x - 1) = [(√6)² - 5)/(√6- 1) = (6 - 5)/(√6 - 1) = 1/(√6 - 1) =
= (√6 +1)/[(√6 -1)(√6 +1) = (√6 + 1)/(6 - 1) = (√6 + 1)/5
zad 3
A = ( - 1 , a ) , B = (2/5 , b ) , C =(c , - 4/3)
f(x) = 4/x
dla punktu A
a = 4/(- 1)
a = - 4/1 = 4
dla punktu B
b = 4 : 2/5 = 4 * 5/2 = 20/2 = 10
dla punktu C
- 4/3 = 4/c
- 4c = 3 * 4
- 4c = 12
4c = - 12
c = - 12/4 = - 3
