Question

Find the domain of each function using interval notation. (Please help I have an exam tomorrow and I’m really stuck)

Answer 1

9.
`(-\infty, 3]`

15.
`(-\infty, -(1)/(2)) \cup (-(1)/(2), \infty)`

Explanation:

9.

The function has a square root. Since you cannot take the square root of a negative number, the expression in the root must be non-negative.

`6 - 2x \ge 0`

`-2x \ge -6`

`x \le 3`

`(-\infty, 3]`

15.

There is a denominator int he function. The denominator cannot equal zero. Set the denominator equal to zero to find out the value that must be excluded from x.

`4x + 2 = 0`

`4x = -2`

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

`(-\infty, -(1)/(2)) \cup (-(1)/(2), \infty)`

Answer 2

For number 9)

Interval notation
`(-\infty,3]`

For number 10)

Interval notation:
`(-\infty,(-1)/(2)) \cup ((-1)/(2),\infty)`.

Explanation:

On square roots you have to make sure the inside is positive or zero.

So the domain of the first one will come from solving

`6-2x \ge 0`

Subtract 6 on both sides:

`-2x \ge -6`

Divide both sides by -2 (flip inequality when divide both sides by negative):

`x \le 3`

The domain is less than or equal to 3.

Interval notation
`(-\infty,3]`

On fractions you have to watch out for dividing by 0.

The domain is all real numbers except when 4x+2=0.

4x+2=0

Subtract 2 on both sides:

4x=-2

Divide both sides by 4:

x=-2/4

Reduce:

x=-1/2

The domain is all real numbers except when x=-1/2

Interval notation:
`(-\infty,(-1)/(2)) \cup ((-1)/(2),\infty)`.