Question

Given f (x) = x2 + 6x + 5 and g(x) = x3 + x2 − 4x − 4, find the domain of f over g of x period

{x ∈ ℝ}
x ≠ −5
x ∈ ℝ
x ≠ −2, 2

Answer 1

`\textsf{Domain}: \quad \{ x \in \mathbb{R}\;|\;x \\eq -1, -2, 2\}`

Explanation:

Given functions:

`\begin{cases} f(x)=x^2+6x+5\\g(x)=x^3+x^2-4x-4\end{cases}`

Factor function f(x):

`\begin{aligned}\implies f(x)&= x^2+6x+5\\&=x^2+x+5x+5\\&=x(x+1)+5(x+1)\\&=(x+5)(x+1)\end{aligned}`

Factor function g(x):

`\begin{aligned}\implies g(x)&=x^3+x^2-4x-4\\&=(x+1)(x^2-4)\\&=(x+1)(x+2)(x-2)\end{aligned}`

Therefore the composite function is:

`\implies (f(x))/(g(x))=((x+5)(x+1))/((x+1)(x+2)(x-2))`

A rational function is undefined when the denominator equals zero.

Therefore, the given composite function is undefined when (x+1)(x+2)(x-2)=0:

• 
  `x+1=0 \implies x=-1`

• 
  `x+2=0 \implies x=-2`

• 
  `x-2=0 \implies x=2`

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

Therefore, as the composite function is undefined when x = -1, x = -2 and x = 2, the domain is:

• 
  `\{ x \in \mathbb{R}\;|\;x \\eq -1, -2, 2\}`

Answer 2

The main limiting issue with the domain this function is potentially picking an x-value that makes that denominator equal 0, because dividing by 0 is bad.

So, the domain will be all real numbers, except for those that make g(x)=0.

g(-1) = 0, g(-2)=0, and g(2) = 0, so -1, -2, and 2 must be excluded from the domain.

Your answer is x ∈ ℝ.