Question

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

Answer 1

Answer:

For number 9)

Interval notation $(-\infty,3]$

For number 10)

Interval notation: $(-\infty,\frac{-1}{2}) \cup (\frac{-1}{2},\infty)$.

Step-by-step 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: $(-\infty,\frac{-1}{2}) \cup (\frac{-1}{2},\infty)$.

Answer 2

Answer:

9. $(-\infty, 3]$

15. $(-\infty, -\dfrac{1}{2}) \cup (-\dfrac{1}{2}, \infty)$

Step-by-step 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.

$4x + 2 = 0$

$4x = -2$

$x = -\dfrac{1}{2}$

$(-\infty, -\dfrac{1}{2}) \cup (-\dfrac{1}{2}, \infty)$