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]}

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