Question:
If m (x) = StartFraction x + 5 Over x minus 1 EndFraction and n(x) = x – 3, which function has the same domain as (m circle n) (x)?
h (x) = StartFraction x + 5 Over 11 EndFraction
h (x) = StartFraction 11 Over x minus 1 EndFraction
h (x) = StartFraction 11 Over x minus 4 EndFraction
h (x) = StartFraction 11 Over x minus 3 EndFraction
Answer:
Option C: [tex]h(x)=\frac{11}{x-4}[/tex] has the same domain as [tex]$(m \circ n)(x)$[/tex]
Explanation:
It is given that [tex]m(x)=\frac{x+5}{x-1}[/tex] and [tex]n(x)=x-3[/tex]
Let us find the domain of [tex]$(m \circ n)(x)$[/tex]
[tex]$\begin{aligned}(m \circ n)(x) &=m(n(x))\\&=m(x-3) \\ &=\frac{(x-3)+5}{(x-3)-1} \\ &=\frac{x+2}{x-4} \end{aligned}$[/tex]
Now, let us equate the denominator equal to zero to determine the domain.
[tex]$x-4=0$[/tex]
[tex]x=4[/tex]
Thus, the function becomes undefined at the point [tex]x=4[/tex]
Hence, the domain of [tex]$(m \circ n)(x)$[/tex] is [tex]$(-\infty, 4) \cup(4, \infty)$[/tex]
Now, we shall find the function which has the same domain as [tex]$(m \circ n)(x)$[/tex]
Option A: [tex]h(x)=\frac{x+5}{11}[/tex]
The function h(x) has the domain of set of all real numbers [tex]$-\infty<x<\infty$[/tex]
Thus, the interval [tex](-\infty,\infty)[/tex] is not the same domain as [tex]$(-\infty, 4) \cup(4, \infty)$[/tex]
Hence, Option A is not the correct answer.
Option B: [tex]h(x)=\frac{11}{x-1}[/tex]
Equating the denominator equal to zero, the function becomes undefined at the point [tex]x=1[/tex]
Thus, the function h(x) has the domain of [tex]$(-\infty,-1) \cup(-1, \infty)$[/tex] is not the same domain as [tex]$(-\infty, 4) \cup(4, \infty)$[/tex]
Hence, Option B is not the correct answer.
Option C: [tex]h(x)=\frac{11}{x-4}[/tex]
Equating the denominator equal to zero, the function becomes undefined at the point [tex]x=4[/tex]
Thus, the function h(x) has the domain of [tex]$(-\infty, 4) \cup(4, \infty)$[/tex] is the same domain as [tex]$(-\infty, 4) \cup(4, \infty)$[/tex]
Hence, Option C is the correct answer.
Option D: [tex]h(x)=\frac{11}{x-3}[/tex]
Equating the denominator equal to zero, the function becomes undefined at the point [tex]x=3[/tex]
Thus, the function h(x) has the domain of [tex]$(-\infty,3) \cup(3, \infty)$[/tex] is not the same domain as [tex]$(-\infty, 4) \cup(4, \infty)$[/tex]
Hence, Option D is not the correct answer.
