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

Answer 1

Eva Abramson

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

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

Consequently,

#x^3+8!=0#

#x^3!=-8#

#x!=-2#

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

To determine the range, follow these steps:

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#

This equation's solution 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 sole limitation 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}#

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

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Answer 2

Kennedy Adams

To find the domain of $f(x) = \frac{x}{x^3 + 8}$ , set the denominator $x^3 + 8$ not equal to zero. Thus, the domain is $x \neq -2$ .

To find the range, observe that as $x$ approaches positive or negative infinity, $f(x)$ approaches zero. The range is all real numbers except zero.

Domain: $x \neq -2$ Range: $f(x)$ is all real numbers except zero.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.