What is the domain and range of #sqrt((5x+6)/2)#?

1 Answer
Costas C.
Mar 15, 2016

Answer:

Domain #x in[-6/5,oo)#
Range #[0,oo)#

Explanation:

You must keep in mind that for the domain:

#sqrt(y)->y>=0#

#ln(y)->y>0#

#1/y->y!=0#

After that, you will be lead to an unequality giving you the domain.

This function is a combination of linear and square functions. Linear has domain #RR#. The square function though must have a positive number inside the square. Therefore:

#(5x+6)/2>=0#

Since 2 is positive:

#5x+6>=0#

#5x>= -6#

Since 5 is positive:

#x>= -6/5#

The domain of the functions is:

#x in[-6/5,oo)#

The range of the root function (outer function) is #[0,oo)# (infinite part can be proven through the limit as #x->oo#).

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