# What is the domain of the function: f(x) =sqrt(x^2(x-3)(x-4))?

Oct 2, 2015

${D}_{f \left(x\right)} = \left(- \infty , 3\right] \cup \left[4 , + \infty\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \sqrt{{x}^{2} \left(x - 3\right) \left(x - 4\right)}$

To find the domain we need to determine which values of $x$ are not valid.

Since the $\sqrt{\text{negative value}}$ is undefined (for Real numbers)

${x}^{2} \left(x - 3\right) \left(x - 4\right) \ge 0$

${x}^{2} \ge 0$ for all $x \in \mathbb{R}$
$\left(x - 3\right) > 0$ for all $x > 3 , \in \mathbb{R}$
$\left(x - 4\right) > 0$ for all $x > 4 , \in \mathbb{R}$

The only combination for which
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} \left(x - 3\right) \left(x - 4\right) < 0$
is when $\left(x - 3\right) > 0$ and $\left(x - 4\right) < 0$

That is the only non-valid values for (Real) $x$ occur when
$\textcolor{w h i t e}{\text{XXX}} x > 3$ and $x < 4$

Oct 2, 2015

$\left(- \infty , 3\right] \cup \left[4 , \infty\right)$

#### Explanation:

The domain is where the radicand (the expression under the square root sign) is non-negative.

We know that ${x}^{2} \ge 0$ for all $x \in \mathbb{R}$.

So in order that ${x}^{2} \left(x - 3\right) \left(x - 4\right) \ge 0$, we must either have ${x}^{2} = 0$ or $\left(x - 3\right) \left(x - 4\right) \ge 0$.

When $x \le 3$, both $\left(x - 3\right) \le 0$ and $\left(x - 4\right) \le 0$, so $\left(x - 3\right) \left(x - 4\right) \ge 0$

When $3 < x < 4$, $\left(x - 3\right) > 0$ and $\left(x - 4\right) < 0$, so $\left(x - 3\right) \left(x - 4\right) < 0$.

When $x \ge 4$, both $\left(x - 3\right) \ge 0$ and $\left(x - 4\right) \ge 0$, so $\left(x - 3\right) \left(x - 4\right) \ge 0$.

So ${x}^{2} \left(x - 3\right) \left(x - 4\right) \ge 0$ when $x \in \left(- \infty , 3\right] \cup \left[4 , \infty\right)$

Note that this domain already includes the point $x = 0$, so the ${x}^{2} = 0$ condition gives us no extra points for the domain.