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)?

Answer 1

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: $h(x)=\frac{11}{x-4}$ has the same domain as $(m \circ n)(x)$

Explanation:

It is given that $m(x)=\frac{x+5}{x-1}$ and $n(x)=x-3$

Let us find the domain of $(m \circ n)(x)$

$\begin{aligned}(m \circ n)(x) &=m(n(x))\\&=m(x-3) \\ &=\frac{(x-3)+5}{(x-3)-1} \\ &=\frac{x+2}{x-4} \end{aligned}$

Now, let us equate the denominator equal to zero to determine the domain.

$x-4=0$

     $x=4$

Thus, the function becomes undefined at the point $x=4$

Hence, the domain of $(m \circ n)(x)$ is $(-\infty, 4) \cup(4, \infty)$

Now, we shall find the function which has the same domain as $(m \circ n)(x)$

Option A: $h(x)=\frac{x+5}{11}$

The function h(x) has the domain of set of all real numbers $-\infty$

Thus, the interval $(-\infty,\infty)$ is not the same domain as $(-\infty, 4) \cup(4, \infty)$

Hence, Option A is not the correct answer.

Option B: $h(x)=\frac{11}{x-1}$

Equating the denominator equal to zero, the function becomes undefined at the point $x=1$

Thus, the function h(x) has the domain of $(-\infty,-1) \cup(-1, \infty)$ is not the same domain as $(-\infty, 4) \cup(4, \infty)$

Hence, Option B is not the correct answer.

Option C: $h(x)=\frac{11}{x-4}$

Equating the denominator equal to zero, the function becomes undefined at the point $x=4$

Thus, the function h(x) has the domain of $(-\infty, 4) \cup(4, \infty)$ is the same domain as $(-\infty, 4) \cup(4, \infty)$

Hence, Option C is the correct answer.

Option D: $h(x)=\frac{11}{x-3}$

Equating the denominator equal to zero, the function becomes undefined at the point $x=3$

Thus, the function h(x) has the domain of $(-\infty,3) \cup(3, \infty)$ is not the same domain as $(-\infty, 4) \cup(4, \infty)$

Hence, Option D is not the correct answer.