y = x²
D(x) = (–∞ ; +∞)
E(y) = [0 ; +∞)
y = –x²
D(x) = (–∞ ; +∞)
E(y) = [0 ; –∞)
y = x² + 1
D(x) = (–∞ ; +∞)
E(y) = [1 ; +∞)
y = x² – 5
D(x) = (–∞ ; +∞)
E(y) = [–5 ; +∞)
1.Գտնել ֆունկցիայի որոշման տիրույթը։
y = 3/x
D(x) = (–∞ ; +∞)\ {0}
y = x – 1/x + 5
D(x) = (–∞ ; +∞)\ {–5}
y = 3/|x|
D(x) = (–∞ ; +∞)\ {0}
y = √x + 4
D(x) = [– 4 ; +∞)
y = √x – 14
D(x) = [14 ; +∞)
y = √x² + 4
D(x) = (–∞ ; +∞)
y = √x² – 25
D(x) = (–∞ ; +∞) \ {–4, –3, –2, –1, 0, 1, 2, 3, 4}
y = x – 1/√x – 6
D(x) = (6 ; +∞)
y = x – 1 /√x² + 6
D(x) = (–∞ ; +∞)
y =|X|/x
D(x) = (–∞ ; +∞) \ {0}
y =|X|/x + 1
D(x) = (–∞ ; +∞) \ {–1}