Using the functions ݂f(x)=-3x and h(x)=(1)/(x+1), find (h ∘ f ((x)). State the domain of ℎ ∘ ݂f in interval notation.

Question

Using the functions ݂
`f(x)=-3x` and
`h(x)=(1)/(x+1)`, find (
`h` ∘
`f` (
`(x)`). State the domain of ℎ ∘ ݂
`f` in interval notation.

Answer 1

`\textsf{Domain:} \quad \left(-\infty, (1)/(3)\right) \cup \left((1)/(3), \infty\right)`

`\textsf{Range:} \quad \left(-\infty, 0\right) \cup \left(0, \infty\right)`

Explanation:

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

The range of a function is the set of all possible output values (y-values) for which the function is defined.

Given functions:

`f(x)=-3x`

`h(x)=(1)/(x+1)`

First, determine the composite function (h o f)(x) by substituting function f(x) in place of the x in function h(x):

`\begin{aligned}(h \circ f)(x)&=h(-3x)\\\\&=(1)/((-3x)+1)\\\\&=(1)/(-3x+1)\end{aligned}`

As function (h o f)(x) is a rational function, it is undefined when its denominator equals zero:

`\begin{aligned}-3x+1&=0\\\\-3x&=-1\\\\x&=(1)/(3)\end{aligned}`

So, the composite function is undefined when x = 1/3.

Therefore, its domain is:

`\left(-\infty, (1)/(3)\right) \cup \left((1)/(3), \infty\right)`

As the degree of the numerator of (h o f)(x) is less than the degree of its denominator, a horizontal asymptote occurs at y = 0. Therefore, the range of the composite function is restricted to:

`\left(-\infty, 0\right) \cup \left(0, \infty\right)`