Question

Find the domain of the function

$f(x) = \sqrt{x^2 - x - 6} + \sqrt{6 - x}$

Answer 1

$(-\infty, -2] \cup [3, 6]$

Answer 2

To find the domain of the function $f(x) = \sqrt{x^2 - x - 6} + \sqrt{6 - x}$, we need to ensure that both terms under the square root are non-negative.

Condition 1: The expression under the first square root must be non-negative.

$x^2 - x - 6 \ge 0$

To solve this quadratic inequality, we first find the roots of the corresponding quadratic equation $x^2 - x - 6 = 0$.

Factoring the quadratic expression:

$(x - 3)(x + 2) \ge 0$

The roots are $x = 3$ and $x = -2$. Since the quadratic has a positive leading coefficient (coefficient of $x^2$ is 1), the parabola opens upwards. Thus, the expression $x^2 - x - 6$ is non-negative when $x$ is less than or equal to the smaller root or greater than or equal to the larger root.

So, $x \le -2$ or $x \ge 3$. In interval notation, this solution set is $D_1 = (-\infty, -2] \cup [3, \infty)$.

Condition 2: The expression under the second square root must be non-negative.

$6 - x \ge 0$

$6 \ge x$

$x \le 6$

In interval notation, this solution set is $D_2 = (-\infty, 6]$.

Finding the Domain:

The domain of $f(x)$ is the intersection of the solution sets from Condition 1 and Condition 2, i.e., $D = D_1 \cap D_2$.

We need to find the values of $x$ that satisfy both conditions:

$x \in ((-\infty, -2] \cup [3, \infty))$ AND $x \in (-\infty, 6]$.

Let's find the intersection of the intervals:

1. Intersection of $(-\infty, -2]$ with $(-\infty, 6]$: This gives $(-\infty, -2]$.
2. Intersection of $[3, \infty)$ with $(-\infty, 6]$: This gives $[3, 6]$.

Combining these two resulting intervals, the domain of $f(x)$ is $(-\infty, -2] \cup [3, 6]$.