Question

The domain of the function $f(x)=\sqrt{x^{2}+2x-15}$ is ________.

• A. $(-\infty,-5) \cup (3,\infty)$
• B. $(-\infty,-5) \cup [3,\infty)$
• C. $(-\infty,-5] \cup (3,\infty)$
• D. $(-\infty,-5] \cup [3,\infty)$
• E. $[-5,3]$

Hint:

Use the wavy curve method for quadratic inequalities.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For a square root function, \(f(x) = \sqrt{g(x)}\), the expression inside the square root, \(g(x)\), must be non-negative (greater than or equal to zero).
Step 2: Key Formula or Approach
To find the domain of \(f(x) = \sqrt{x^2 + 2x - 15}\), we must solve the inequality:
\[ x^2 + 2x - 15 \geq 0 \] Step 3: Detailed Explanation
1. Factor the quadratic expression.
We need to find two numbers that multiply to -15 and add to +2. These numbers are +5 and -3.
\[ (x + 5)(x - 3) \geq 0 \] 2. Find the critical points.
The critical points are the values of x where the expression equals zero.
\(x + 5 = 0 \Rightarrow x = -5\)
\(x - 3 = 0 \Rightarrow x = 3\)
These points divide the number line into three intervals: \((-\infty, -5)\), \((-5, 3)\), and \((3, \infty)\).
3. Test the intervals.
We can test a value from each interval to see if the inequality \((x + 5)(x - 3) \geq 0\) holds true.
- Interval 1: \((-\infty, -5)\)
Let's pick \(x = -6\).
\((-6 + 5)(-6 - 3) = (-1)(-9) = 9\). Since \(9 \geq 0\), this interval is part of the domain.
- Interval 2: \((-5, 3)\)
Let's pick \(x = 0\).
\((0 + 5)(0 - 3) = (5)(-3) = -15\). Since \(-15 \not\geq 0\), this interval is not part of the domain.
- Interval 3: \((3, \infty)\)
Let's pick \(x = 4\).
\((4 + 5)(4 - 3) = (9)(1) = 9\). Since \(9 \geq 0\), this interval is part of the domain.
4. Consider the endpoints.
The inequality is \(\geq\) (greater than or equal to), which means the critical points themselves (\(x = -5\) and \(x = 3\)) are included in the domain. We use square brackets \([]\) to indicate inclusion.
5. Combine the results.
The domain consists of the first interval, the third interval, and the endpoints.
Domain = \((-\infty, -5] \cup [3, \infty)\).
Step 4: Final Answer
The domain of the function is \((-\infty, -5] \cup [3, \infty)\).