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

Question

Genesis Allen · Accepted Answer

Domain = #RR-{-2}#
Range = #RR# To identify potential discontinuities, factorize the denominator first as the sum of two cubes, and then as a trinomial to obtain #x/((x+2)(x^2-2x+4)# With no real roots, the trinomial in the denominator is an irreducible quadratic. Hence the only vertical asymptote is at #x=-2# The horizontal asymptote would usually occur at #lim_(x-oo)f(x) =0#, since the denominator dominates the numerator and increases faster. However in this case, 0 is in the range of the function since x = 0 in the numerator would set the output value y to 0. Thus, the range is all real numbers y, and the domain is all real numbers x, minus -2. Graphically: [-10, 10, -5, 5]} graph{x/(x^3+8)

Kennedy Adams · Answer

undefined To find the domain, set the denominator $ x^3 + 8 $ not equal to zero and solve for $ x $. The range can be found by analyzing the behavior of the function as $ x $ approaches positive and negative infinity.

Domain: $ x \neq -2 $

Range: $ f(x) $ approaches 0 as $ x $ approaches positive or negative infinity. Therefore, the range is all real numbers except 0.

Jace Anderson · Answer

undefined To find the domain of $ f(x) = \frac{x}{x^3 + 8} $, we need to identify any values of $ x $ that would make the denominator zero, as division by zero is undefined.

The denominator $ x^3 + 8 $ will be zero when $ x = -2 $. Therefore, the domain of $ f(x) $ is all real numbers except $ x = -2 $.

To find the range, we need to consider the behavior of the function as $ x $ approaches positive and negative infinity.

As $ x $ approaches positive infinity, the function approaches zero. As $ x $ approaches negative infinity, the function also approaches zero.

Thus, the range of $ f(x) $ is all real numbers, except zero.