Question

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

• A. $(-\infty,-2)\cup[1,\infty)$
• B. $(-\infty,-2]\cup(1,\infty)$
• C. $(-\infty,-2)\cup(1,\infty)$
• D. $(-\infty,-2]\cup[1,\infty)$
• E. $(-\infty,1)\cup[0,\infty)$

Hint:

Use the wavy curve method to solve 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 because the square root of a negative number is not a real number.
Step 2: Key Formula or Approach:
To find the domain of \( f(x) = \sqrt{x^2 + x - 2} \), we must solve the inequality:
\[ x^2 + x - 2 \geq 0 \] Step 3: Detailed Explanation:
First, we find the roots of the quadratic equation \( x^2 + x - 2 = 0 \) by factoring.
\[ x^2 + 2x - x - 2 = 0 \] \[ x(x + 2) - 1(x + 2) = 0 \] \[ (x + 2)(x - 1) = 0 \] The roots are \( x = -2 \) and \( x = 1 \). These are the critical points that divide the number line into three intervals: \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \).
Now, we test a value from each interval to see where the inequality \( (x + 2)(x - 1) \geq 0 \) holds true.
Interval 1: \( (-\infty, -2) \)
Let's pick \( x = -3 \).
\( (-3 + 2)(-3 - 1) = (-1)(-4) = 4 \). Since \( 4 \geq 0 \), this interval is part of the domain.
Interval 2: \( (-2, 1) \)
Let's pick \( x = 0 \).
\( (0 + 2)(0 - 1) = (2)(-1) = -2 \). Since \( -2<0 \), this interval is not part of the domain.
Interval 3: \( (1, \infty) \)
Let's pick \( x = 2 \).
\( (2 + 2)(2 - 1) = (4)(1) = 4 \). Since \( 4 \geq 0 \), this interval is part of the domain.
Since the inequality is \( \geq 0 \), the roots themselves (\( x = -2 \) and \( x = 1 \)) are included in the domain.
Combining the valid intervals, the domain is \( x \leq -2 \) or \( x \geq 1 \).
Step 4: Final Answer:
In interval notation, the domain is \( (-\infty, -2] \cup [1, \infty) \).