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What is the domain and range of #y= (sqrt (x+4))/x#?

1 Answer

Nov 3, 2015

#x\in[-4,0)uu(0,oo)#
#yin(-oo,oo)#

Explanation:

#x# cannot be less than -4 due to square root of a negative number.
#x# cannot be zero due to division by zero.

When #-4<=x<0#, #-oo<\y<=0#.
When #0<\x<\oo#, #0<\y<\oo#.

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