How do you find the domain and range of #f(x)=x/(x^3+8) #?

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Answer 1 · 2018-01-23T11:12:46

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

As you cannot divide by #0#, the denominator is #!=0#

Therefore,

#x^3+8!=0#

#x^3!=-8#

#x!=-2#

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

To calculate the range, proceed as follows

Let #y=x/(x^3+8)#

#y(x^3+8)=x#

#x^3y+8y- x=0#

#x^3y-x+8y=0#

This is a cubic equation in #x#

The solution of this equation is

#x=(-4+(((-1/(3y))^2)+(-1/(3y))^3)^(1/2))^(1/3)+(-4-((-1/(3y))^2)+(-1/(3y))^3)^(1/2)))^(1/3)#

The only restriction is

#((-1/(3y))^2+(-1/(3y))^3)>=0#

#1/(9y^2)-1/(27y^3)>=0#

#(3y-1)/(27y^3)>=0#

So #y!=0#

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

graph{x/(x^3+8) [-7.9, 7.904, -3.95, 3.95]}