# Question #56413

Aug 29, 2015

For part (a): Domain: $\left(- \infty , - 3\right]$
For part (b): Domain: $\left(- \infty , + \infty\right)$
For part (c): Domain: $\left[2 , 4\right]$

#### Explanation:

In all three cases, you're dealing with functions that equal the square root of an expression.

Right from the start, you need to take into account the fact that, for real numbers, you cannot take the square root of a negative number.

This means that you need to check these expressions for any value of $x$ that would make them negative, and exclude these values from the functions' domain.

$f \left(x\right) = \sqrt{- 3 x - 9}$

You need to have

$\left(- 3 x - 9\right) \ge 0$

$- 3 x \ge 9 \implies x \le \frac{9}{\left(- 3\right)} = - 3$

The domain of the function will thus include any value of $x$ that is greater than or equal to $\left(- 3\right)$. Any other value will be excluded. This means that the function's domain will be $\left(- \infty , - 3\right]$.

graph{sqrt(-3x-9) [-14.24, 14.24, -7.12, 7.12]}

Do the same for the second one.

$g \left(x\right) = \sqrt{{x}^{2} + 4}$

Now, because ${x}^{2} \ge 0$ for all real values of $x$, the expression under the square root will never be positive. In fact, it will never be smaller than $4$, since that's the value you'd get for $x = 0$.

This means that you don't have any restrictions for the function's domain, which will be $x \in \mathbb{R}$, or $x \in \left(- \infty , + \infty\right)$.

graph{sqrt(x^2 + 4) [-41.1, 41.14, -20.53, 20.56]}

Now for the third one.

$h \left(x\right) = \sqrt{2 x - 4} + \sqrt{- 5 x + 20}$

You need the expressions that are under the square roots to be valid at the same time. You will thus have

$2 x - 4 \ge 0$

$2 x \ge 4 \implies x \ge 2$

and

$- 5 x + 20 \ge 0$

$- 5 x \ge - 20 \implies x \le 4$

The domain of the function will thus be $\left[2 , 4\right]$, since any value of $x$ outside this interval will make of the the two square roots undefined.

graph{sqrt(2x-4) + sqrt(-5x + 20) [-16, 16.05, -8, 8.02]}