What is the domain and range of #h(x)= 10/(x^2-2x)#?

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Answer 1 · 2016-06-19T07:04:26

Domain is #(-oo,0) uu (0,2)uu(2,+oo)#
Range is #(-oo,-40/9]uu(0,+oo)#

The domain is obtained by solving:

#x^2-2x!=0#

#x(x-2)!=0#

#x!=0 and x!=2#

You can find the range by calculating the inverse function

Let y=h(x)

so

#y=10/(x^2-3x)#

#yx^2-3xy-10=0#

#x=(3y+-sqrt(9y^2-4y(-10)))/(2y)#

you can find its domain by solving:

#9y^2+40y>=0 and y!=0#

#y(9y+40)>=0 and y!=0#

#y<=-40/9 or y>0#