\(\displaystyle x\) cannot equal a value which would make the denominator equal to 0.  In order to figure out what that value is, we must first simplify this fraction, then set each factor of the denominator equal to 0.  As follows:

\(\displaystyle \frac{x^{2}-4}{x^2-x-2}\)

First, simplify by finding the original binomial multiples:

\(\displaystyle \frac{(x-2)(x+2)}{(x-2)(x+1)}\)

Now, set each factor of the denominator equal to 0

\(\displaystyle x+1=0\)

\(\displaystyle x=-1\)

\(\displaystyle x-2=0\)

\(\displaystyle x=2\)

If \(\displaystyle x=-1\) OR \(\displaystyle x=2\), the denominator will equal 0.  \(\displaystyle -1\) is the only choice provided by the answers.

Note: Even though you can cancel out \(\displaystyle x-2\) from the numerator and denominator, \(\displaystyle x\) still cannot be equal to two. The graph would have a hole at \(\displaystyle x=2\) and an aymptote at \(\displaystyle x=-1\).

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:

\(\displaystyle 3x-2 = 0\)

\(\displaystyle 3x = 2\)

\(\displaystyle x= \frac{2}{3}\)

Therefore, the domain of \(\displaystyle f\) is the set of all real numbers except \(\displaystyle \frac{2}{3}\) - that is, 

\(\displaystyle \left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )\)

The domain of \(\displaystyle g\) is restricted by two different denominators, neither of which can be equal to 0, so the excluded values are:

\(\displaystyle x-2 = 0 \Rightarrow x = 2\)

\(\displaystyle x-3 = 0 \Rightarrow x =3\)

The correct response is therefore \(\displaystyle \left ( -\infty, 2 \right ) \cup (2,3) \cup (3, \infty)\).

The numerator, being a polynomial, does not restrict our domain. The denominator, however, does restrict it to the values for which it is not equal to 0. We set the denominator equal to 0 to find the excluded values:

\(\displaystyle x^{2} - 9 = 0\)

\(\displaystyle (x-3) (x+3) = 0\)

\(\displaystyle x- 3 = 0 \Rightarrow x = 3\)

\(\displaystyle x+3 = 0 \Rightarrow x = -3\)

The domain, in interval notation, is therefore

\(\displaystyle \left ( -\infty, -3 \right ) \cup (-3,3) \cup (3, \infty)\).

The domain of \(\displaystyle g\) is restricted by two things.

First, the expression within the radical in the numerator must be nonnegative. We therefore solve for \(\displaystyle x\) in the inequality

\(\displaystyle x + 3 \geq 0\)

\(\displaystyle x \geq -3\),

or, in interval notation, \(\displaystyle \left [-3, \infty \right )\)

Second, the expression in the denominator must be nonzero.  Therefore, we set the denominator equal to zero to determine the excluded value(s):

\(\displaystyle x - 4 = 0\)

\(\displaystyle x = 4\)

We exclude 4 from \(\displaystyle \left [-3, \infty \right )\), so the correct response is 

\(\displaystyle \left [-3, 4\right ) \cup \left (4, \infty \right )\)

There are two things restricting the domain of \(\displaystyle h\).

One is the radical symbol in the numerator. The expression inside the radical must be nonnegative, so solve the inequality:

\(\displaystyle x - 1 \geq 0\)

\(\displaystyle x \geq 1\),

or, in interval notation, \(\displaystyle [1, \infty)\)

The other is the denominator, which must be equal to 0, so set, and solve for \(\displaystyle x\) in, the equation:

\(\displaystyle x^{2}- 3x - 4 = 0\)

\(\displaystyle (x-4)(x+1) = 0\)

\(\displaystyle x - 4 = 0 \Rightarrow x = 4\)

\(\displaystyle x+ 1 = 0 \Rightarrow x = -1\)

\(\displaystyle -1\) is already excluded from the domain; we exclude 4, so the domain is

\(\displaystyle \left [1, 4 \right ) \cup \left ( 4 ,\infty)\).

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the expression in the denominator, which must be nonzero. Furthermore, the radicand must be nonnegative. Combined, these facts mean that the radicand must be positive, and that the following inequality be solved:

\(\displaystyle 10 - x > 0\)

\(\displaystyle 10 > x\) or, equivalently, \(\displaystyle x < 10\).

In interval form, this is \(\displaystyle (- \infty, 10 )\)

The domain of the product of two functions is the intersection of the domains of the individual functions.

The domain of \(\displaystyle f\) is restricted to all values of \(\displaystyle x\) that yield a nonzero denominator. Since this means that 

\(\displaystyle x- 4 \ne 0\),

then, subsequently,

\(\displaystyle x \ne 4\),

so the domain is the set of all real numbers except 4.

Similarly, the domain of \(\displaystyle g\) is restricted to all values of \(\displaystyle x\) that yield a nonzero denominator. This set is found to be the set of all real numbers except 7.

The intersection of these sets is the set of all real numbers except 4 and 7, or

\(\displaystyle \left (-\infty, 4 \right ) \cup \left ( 4,7\right ) \cup \left ( 7, \infty \right )\).

The definition of \(\displaystyle f\) has two denominators of two fractions (one within the other), so we must exclude the values of \(\displaystyle x\) that make either denominator equal to zero.

One denominator is \(\displaystyle x-2\). Since it cannot be zero, we have 

\(\displaystyle x- 2 \ne 0\),

and, subsequently,

\(\displaystyle x \ne 2\).

The value 2 is excluded from the domain.

The other denominator is \(\displaystyle 4- \frac{2}{x-2}\). If it is equal to zero, then

\(\displaystyle 4- \frac{2}{x-2} = 0\)

\(\displaystyle 4= \frac{2}{x-2}\)

\(\displaystyle 4\cdot \frac{x-2}{4} = \frac{2}{x-2} \cdot \frac{x-2}{4}\)

\(\displaystyle x-2 = \frac{1}{2}\)

\(\displaystyle x=2 \frac{1}{2}\)

Therefore, this value is also excluded from the domain.

The correct domain is the set \(\displaystyle \left ( -\infty , 2 \right ) \cup \left ( 2, 2\frac{1}{2}\right ) \cup \left ( 2\frac{1}{2}, \infty \right )\).

The numerator \(\displaystyle x-4\), being a polynomial, does not restrict the domain.

\(\displaystyle x\) appears as a radicand of a square root in the definition of \(\displaystyle f\), so \(\displaystyle x\) must be restricted to nonnegative numbers. Also, \(\displaystyle \sqrt{x}\) is a denominator, which means 0 must also be excluded. Therefore, we are restricted so far to positive numbers.

There is one more denominator, which is \(\displaystyle 1- \frac{3}{\sqrt x}\); it must be nonzero. We set this equal to zero to determine any additional value(s) that must be excluded:

\(\displaystyle 1- \frac{3}{\sqrt x} = 0\)

\(\displaystyle 1= \frac{3}{\sqrt x}\)

\(\displaystyle 1 \cdot \sqrt{x}= \frac{3}{\sqrt x} \cdot \sqrt{x}\)

\(\displaystyle \sqrt{x} = 3\)

\(\displaystyle x= 9\)

Therefore, the domain is the set of all positive numbers except 9 - or

\(\displaystyle \left ( 0, 9\right ) \cup \left ( 9, \infty \right )\).