Answer:
3. Last choice: Domain: (-3, 0) U (0, ∞)
4. Last choice: Domain: (-∞, -3) U (-3, ∞), Range: (-∞, 0) U (0, ∞)
Step-by-step explanation:
3.
[tex](\frac{f}{g})(x) = \frac{1}{x}/ \sqrt{x+3}[/tex]
=
[tex](\frac{f}{g})(x) = \frac{1}{x\sqrt{x+3}}[/tex]
This function will be undefined at x = 0 and x = -3 because the denominator would be equal to 0. This means that the function has vertical asymptotes at x = 0 and x = -3, therefore:
Domain: (-3, 0) U (0, ∞)
*** Remember, x < -3 is not included in the domain because a square root of a negative number does not exist.***
4.
[tex]f(x) = \frac{1}{2x} - 3[/tex]
Use "y" instead of f(x), and swap positions of x and y:
[tex]x = \frac{1}{2y} - 3[/tex]
Simplify to isolate for "y":
[tex]x+3 = \frac{1}{2y}[/tex]
[tex]2y(x+3) = 1[/tex]
[tex]2y = \frac{1}{x+3}[/tex]
[tex]y = \frac{1}{2(x+3)}[/tex]
Based on the denominator, the function has a vertical asymptote at x = -3, therefore:
Domain: (-∞, -3) U (-3, ∞)
There is also an EBA asymptote of y = 0 since the denominator has a higher degree than the numerator, so:
Range: (-∞, 0) U (0, ∞)
