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The answer is e. To determine the domain you must first find all possible x-values that allow a function to work resulting in real y-values. The bottom number of a fraction can never be 0. In all other options 3 is one of the real numbers.

Answer:

b) [tex]\frac{3}{x-3}[/tex]

e) [tex]\frac{1}{\sqrt{x-3}}[/tex]

Step-by-step explanation:

Since,  f(x) = 3x + 3, f(x) = 3x² and f(x) = 3x - 3 are polynomial function,

We know that,

The domain of a polynomial is the set of all real numbers,

⇒ Thus, the domain of the functions f(x) = 3x + 3, f(x) = 3x² and f(x) = 3x - 3 is the set of all real numbers,

Now, [tex]\frac{3}{x-3}[/tex] and [tex]\frac{1}{\sqrt{x-3}}[/tex] are rational functions,

We know that,

The domain of a rational function is the set of all real numbers except those for which denominator = 0,

[tex]x-3=0\implies x = 3[/tex]

Hence, the domain of functions  [tex]\frac{3}{x-3}[/tex] and [tex]\frac{1}{\sqrt{x-3}}[/tex] is the set of all real numbers except x = 3.

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