Question

The set of all x for which none of the functions is defined $f\left( x \right) = {\log _2}\left( {\dfrac{{x - 1}}{{x + 3}}} \right)$ and $g\left( x \right) = \sqrt {\dfrac{1}{{{x^2} - 9}}}$ is:
A) $\left( { - 3,1} \right)$
B) $\left( { - 3,2} \right)$
C) $( - 3,1]$
D) $\left[ { - 3,2} \right]$

Answer 1

We first find the domains of both the given functions. Then, we figure out the intervals of x for which the functions are not defined. We find the intersection of the intervals where both the functions are not defined and get to the required answer. For finding the domains, we must know some basic mathematical rules such as the entity under the square root should be positive.

Complete step by step solution:
We have, $f\left( x \right) = {\log _2}\left( {\dfrac{{x - 1}}{{x + 3}}} \right)$.
We know that we cannot input negative entities into the logarithmic functions.
So, we get, $\left( {\dfrac{{x - 1}}{{x + 3}}} \right) > 0$.
So, either both numerator and denominator should be negative or both should be positive.
Either $\left( {x - 1} \right) > 0$ and $\left( {x + 3} \right) > 0$
$\Rightarrow x > 1$ or $\Rightarrow x > - 3$
Combining the above two inequalities, we can say $x > 1$, so
$x \in (1, \infty)$
Or $\left( {x - 1} \right) < 0$ and $\left( {x + 3} \right) < 0$
$\Rightarrow x < 1$ and $\Rightarrow x < \- 3$
Combining the above two inequalities, we can say, $x < \- 3$. So
$x \in (-\infty, -3)$
Hence, we have the domain of the function $f\left( x \right) = {\log _2}\left( {\dfrac{{x - 1}}{{x + 3}}} \right)$ as $x \in \left( { - \infty , - 3} \right) \cup \left( {1,\infty } \right)$. -- (1)

Now, for $g\left( x \right) = \sqrt {\dfrac{1}{{{x^2} - 9}}}$, we know that negative values are not permitted in square root. Also, since the square root is present in the denominator, so it should not be zero as well.
So, we have, ${x^2} - 9 > 0$.
$\Rightarrow \left( {x - 3} \right)\left( {x + 3} \right) > 0$
Now, either both the terms must be positive, or both must be negative.
So, we have,
Either $\left( {x - 3} \right) > 0$ and $\left( {x + 3} \right) > 0$
$\Rightarrow x > 3$ and $\Rightarrow x > - 3$
So, taking the intersection of both inequalities, $x > 3$.
Or $\left( {x - 3} \right) < 0$ and $\left( {x + 3} \right) < 0$
$\Rightarrow x < 3$ and $\Rightarrow x < \- 3$
So, taking the intersection, we have, $x < \- 3$.
So, domain for $g\left( x \right) = \sqrt {\dfrac{1}{{{x^2} - 9}}}$ is $x \in \left( { - \infty , - 3} \right) \cup \left( {3,\infty } \right)$.
So, the function $g\left( x \right) = \sqrt {\dfrac{1}{{{x^2} - 9}}}$ is not defined in the interval $x \in \left( { - 3,3} \right)$ and function $f\left( x \right) = {\log _2}\left( {\dfrac{{x - 1}}{{x + 3}}} \right)$ is not defined in interval $x \in \left( { - 3,1} \right)$. -- (2)

Taking the intersection of the intervals in which both functions (1) and (2) are not defined, we get, $x \in \left( { - 3,1} \right)$.
Therefore, option (A) is the correct answer.

Note:
We must remember that we cannot input negative values along with zero into the logarithmic function. The open intervals means that the endpoints are not included in the interval and these are represented by rounded brackets. Closed intervals include the endpoints as well and are represented by square brackets.