Question

Please help, I have been struggling for a whole week (40pts)

Answer 1

3. Last choice: Domain: (-3, 0) U (0, ∞)

4. Last choice: Domain: (-∞, -3) U (-3, ∞), Range: (-∞, 0) U (0, ∞)

Explanation:

3.

`((f)/(g))(x) = (1)/(x)/ √(x+3)`

=

`((f)/(g))(x) = (1)/(x√(x+3))`

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.

`f(x) = (1)/(2x) - 3`

Use "y" instead of f(x), and swap positions of x and y:

`x = (1)/(2y) - 3`

Simplify to isolate for "y":

`x+3 = (1)/(2y)`

`2y(x+3) = 1`

`2y = (1)/(x+3)`

`y = (1)/(2(x+3))`

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, ∞)