The function can be rewritten as follows:

\(\displaystyle f(x)= \frac{x- 4}{x-7}\)

\(\displaystyle f(x)= \frac{x- 7 + 3}{x-7}\)

\(\displaystyle f(x)= \frac{x- 7 }{x-7}+ \frac{ 3}{x-7}\)

\(\displaystyle f(x)=1+ \frac{ 3}{x-7}\)

The expression \(\displaystyle \frac{ 3}{x-7}\) can assume any value except for 0, so the expression \(\displaystyle f(x)=1+ \frac{ 3}{x-7}\) can assume any value except for 1. The range is therefore the set of all real numbers except for 1, or

\(\displaystyle (-\infty, 1) \cup (1, \infty)\).

This choice is not among the responses.

In a rational function, the domain excludes exactly the value(s) of the variable which make the denominator equal to 0. Set the denominator to find these values:

\(\displaystyle x- 7 =0\)

\(\displaystyle x = 7\)

The domain is the set of all real numbers except 7 - that is, \(\displaystyle (-\infty, 7) \cup (7, \infty)\).

Every real number has one real cube root, so there are no restrictions on the radicand of a cube root expression. The domain is the set of all real numbers.

\(\displaystyle -1 \leq \cos x \leq 1\) for any real value of \(\displaystyle x\).

Therefore,

\(\displaystyle -4(-1) \cdot \geq -4 \cdot \cos x \geq -4 \cdot 1\)

\(\displaystyle 4 \geq -4 \cos x \geq -4\)

\(\displaystyle 4 + 2 \geq -4 \cos x + 2 \geq -4 + 2\)

\(\displaystyle 6 \geq 2-4 \cos x \geq -2\)

\(\displaystyle -2 \leq f(x) \leq 6\)

The range is \(\displaystyle [-2,6]\).

\(\displaystyle -1 \leq \cos \theta \leq 1\) for any real value of \(\displaystyle \theta\), so

\(\displaystyle -1 \leq \cos 4x \leq 1\)

\(\displaystyle -1 \cdot \left (-1 \right ) \geq -1 \cdot \cos 4x \geq -1 \cdot 1\)

\(\displaystyle 1 \geq - \cos 4x \geq -1\)

\(\displaystyle 1+ 2 \geq - \cos 4x + 2 \geq -1 + 2\)

\(\displaystyle 3 \geq 2 - \cos 4x \geq 1\)

\(\displaystyle 1 \leq f(x) \leq 3\),

making the range \(\displaystyle \left [ 1,3\right ]\).

The radicand within a square root symbol must be nonnegative, so

\(\displaystyle 16 - x^{2} \geq 0\)

\(\displaystyle 16 \geq x^{2}\)

\(\displaystyle x^{2} \leq 16\)

This happens if and only if \(\displaystyle -4 \leq x \leq 4\), so the domain of \(\displaystyle f\) is \(\displaystyle [-4,4]\).

\(\displaystyle f (x) = \sqrt{16-x^{2}}\) assumes its greatest value when \(\displaystyle 16-x^{2}\), which is the point on \(\displaystyle [-4,4]\) where \(\displaystyle x^{2}\) is least - this is at \(\displaystyle x = 0\).

\(\displaystyle f (0) = \sqrt{16-0^{2}} = \sqrt{16} = 4\)

Similarly, \(\displaystyle f (x) = \sqrt{16-x^{2}}\) assumes its least value when \(\displaystyle 16-x^{2}\), which is the point on \(\displaystyle [-4,4]\) where \(\displaystyle x^{2}\) is greatest - this is at \(\displaystyle x = \pm 4\).

\(\displaystyle f (4) = \sqrt{16-4^{2}} = \sqrt{0} = 0\)

\(\displaystyle f (-4) = \sqrt{16- \left (-4 \right )^{2}} = \sqrt{0} = 0\)

Therefore, the range of \(\displaystyle f\) is \(\displaystyle [0, 4]\).

\(\displaystyle f(x) = \frac{5}{\sin 4x}\) can be rewritten as \(\displaystyle f (x) = 5 \csc 4x\).

For all real values of \(\displaystyle \theta\),

\(\displaystyle \csc \theta\leq -1\) or \(\displaystyle \csc \theta\geq 1\).

Therefore,

\(\displaystyle \csc 4x \leq -1\) or \(\displaystyle \csc 4x\geq 1\) and

\(\displaystyle 5 \csc 4x \leq -5\) or \(\displaystyle 5 \csc 4x\geq 5\) .

The range of \(\displaystyle f\) is \(\displaystyle (- \infty, -5] \cup [5, \infty)\).

The domain of a function is all the x-values that in that function. The function \(\displaystyle f(x)=x^2+3\) is a upward facing parabola with a vertex as (0,3). The parabola keeps getting wider and is not bounded by any x-values so it will continue forever. Parenthesis are used because infinity is not a definable number and so it can not be included.

Notice this function resembles the parent function \(\displaystyle y=\sqrt {x}\).  The value of \(\displaystyle x\) must be zero or greater.

Set up an inequality to determine the domain of \(\displaystyle x\).

\(\displaystyle -10x+3\geq 0\)

Subtract three from both sides.

\(\displaystyle -10x+3-3\geq 0-3\)

\(\displaystyle -10x\geq -3\)

Divide by negative ten on both sides.  The sign will switch.

\(\displaystyle \frac{-10x}{-10}\geq \frac{-3}{-10}\)

The domain is:  \(\displaystyle x\leq \frac{3}{10}\)

Start by considering the term \(\displaystyle \frac{2}{x}\). \(\displaystyle \frac{2}{x}\) will hold for all values of \(\displaystyle x\), except when \(\displaystyle x=0\). Thus, \(\displaystyle \frac{2}{x}-3\) must be defined by all values except \(\displaystyle -3\) since the equation is just shifted down by \(\displaystyle 3\).