Ano ang domain at saklaw ng y = (2x ^ 2) / (x ^ 2 - 1)?

Sagot:

Ang domain ay #x sa (-oo, -1) uu (-1,1) uu (1, + oo) #

Ang hanay ay #y in (-oo, 0 uu (2, + oo) #

Paliwanag:

Ang pag-andar ay

# y = (2x ^ 2) / (x ^ 2-1) #

Pinasisiya namin ang denamineytor

# y = (2x ^ 2) / ((x + 1) (x-1)) #

Samakatuwid, #x! = 1 # at #x! = - 1 #

Ang domain ng y ay #x sa (-oo, -1) uu (-1,1) uu (1, + oo) #

Hayaan ang rearrage ang function

#y (x ^ 2-1) = 2x ^ 2 #

# yx ^ 2-y = 2x ^ 2 #

# yx ^ 2-2x ^ 2 = y #

# x ^ 2 = y / (y-2) #

# x = sqrt (y / (y-2)) #

Para sa # x # sa isang solusyon, # y / (y-2)> = 0 #

Hayaan #f (y) = y / (y-2) #

Kailangan namin ang isang sign na tsart

#color (white) (aaaa) ## y ##color (white) (aaaa) ## -oo ##color (white) (aaaaaa) ##0##color (white) (aaaaaaa) ##2##color (white) (aaaa) ## + oo #

#color (white) (aaaa) ## y ##color (white) (aaaaaaaa) ##-##color (white) (aaa) ##0##color (white) (aaa) ##+##color (white) (aaaa) ##+#

#color (white) (aaaa) ## y-2 ##color (white) (aaaaa) ##-##color (white) (aaa) ##color (white) (aaa) ##-##color (white) (aa) ##||##color (white) (aa) ##+#

#color (white) (aaaa) ##f (y) ##color (white) (aaaaaa) ##+##color (white) (aaa) ##0##color (white) (aa) ##-##color (white) (aa) ##||##color (white) (aa) ##+#

Samakatuwid, #f (y)> = 0 # kailan #y in (-oo, 0 uu (2, + oo) #

graph {2 (x ^ 2) / (x ^ 2-1) -16.02, 16.02, -8.01, 8.01}