Odpowiedź:
y = 1/(x - q) - 3 , x₀ = 2
założenie:
x ≠ q
1/(2 - q) - 3 = 0 | * (2 - q)
1 - 3(2 - q) = 0
1 - 6 + 3q = 0
- 5 + 3q = 0
3q = 5
q = 5/3 = 1 2/3
Df: x ∈ R \ { 1 2/3 }
[tex]y=\frac{1}{x-q} -3\\0=\frac{1}{2-q} -3\\q\epsilon R/\{2\}\\\-\frac{1}{2-q} =-3\\1=6-3q\\3q=6-1\\3q=5|:3\\q=1\frac{2}{3}[/tex]
