What is the domain and range of #(x^2+2)/(x+4)#?

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Answer 1 · 2018-04-06T19:38:52

The domain is #x in RR-{-4}#. The range is #y in (-oo, -16.485] uu [0.485, +oo)#

The denominator is #!=0#

#x+4!=0#

#x!=-4#

The domain is #x in RR-{-4}#

To find the range, proceed as follws

Let #y=(x^2+2)/(x+4)#

#y(x+4)=x^2+2#

#x^2-yx+2-4y=0#

This is a quadratic equation in #x^2# and in order to have solutions

the discriminant #Delta>=0#

Therefore

#Delta=(-y)^2-4(1)(2-4y)>=0#

#y^2-16y-8>=0#

The solutions are

#y=(-16+-sqrt((-16)^2-4(1)(-8)))/2=(-16+-16.97)/2#

#y_1=-16.485#

#y_2=0.485#

The range is #y in (-oo, -16.485] uu [0.485, +oo)#

graph{(x^2+2)/(x+4) [-63.34, 53.7, -30.65, 27.85]}