How do you find the domain and range of #f(x) = (2x+1) / (2x^2 + 5x + 2)#?

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Answer 1 · 2017-09-08T08:38:40

The domain is #x in RR//{-2}#
The range is #y in RR//{0}#

The denominator must de #!=0#

So,

#2x^2+5x+2 = (2x+1)(x+2)!=0#

#f(x)=(2x+1)/((2x+1)(x+2))=1/(x+2)#

Therefore,

The domain is #x in RR//{-2}#

To find the range, we proceed as follows

Let #y=(2x+1)/(2x^2+5x+2 )=1/(x+2)#

#y(x+2 )=1#

#yx+2y=1#

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

The denominator must be #!=0#

Therefore,

The range is #y in RR//{0}#

graph{(2x+1)/(2x^2+5x+2) [-18.02, 18.03, -9.01, 9.01]}