Answer:
Part 1) [tex]f(x)=\frac{x+4}{x}[/tex] -----> [tex]x\neq 0[/tex]
Part 2) [tex]f(x)=\frac{x}{x+4}[/tex] ----> [tex]x\neq -4[/tex]
Part 3) [tex]f(x)=x(x+4)[/tex] ----> All real numbers
Part 4) [tex]f(x)=\frac{4}{x^2+8x+16}[/tex] ----> [tex]x\neq -4[/tex]
Step-by-step explanation:
we know that
The domain of a function is the set of all possible values of x
Part 1) we have
[tex]f(x)=\frac{x+4}{x}[/tex]
we know that
In a quotient the denominator cannot be equal to zero
so
For the value of x=0 the function is not defined
therefore
The domain is
[tex]x\neq 0[/tex]
Part 2) we have
[tex]f(x)=\frac{x}{x+4}[/tex]
we know that
In a quotient the denominator cannot be equal to zero
so
For the value of x=-4 the function is not defined
therefore
The domain is
[tex]x\neq -4[/tex]
Part 3) we have
[tex]f(x)=x(x+4)[/tex]
Applying the distributive property
[tex]f*(x)=x^2+4x[/tex]
This is a vertical parabola open upward
The function is defined by all the values of x
therefore
The domain is all real numbers
Part 4) we have
[tex]f(x)=\frac{4}{x^2+8x+16}[/tex]
we know that
In a quotient the denominator cannot be equal to zero
so
Equate the denominator to zero
[tex]x^2+8x+16=0[/tex]
Remember that
[tex]x^2+8x+16=(x+4)^2[/tex]
([tex]x+4)^2=0[/tex]
The solution is x=-4
so
For the value of x=-4 the function is not defined
therefore
The domain is
[tex]x\neq -4[/tex]
