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Chapter 0: Problem 20

Find the domain of the function. $$ F(x)=\sqrt{x^{2}-2 x-3} $$

Short Answer

Expert verified
The domain of the function \(F(x)=\sqrt{x^{2}-2 x-3}\) is \(x \in (-\infty, -1] \cup [3, \infty)\).

Step by step solution

01

Identify the inequality to solve

The expression under the square root must be greater than or equal to zero for the function to be defined, so we need to solve the inequality \(x^2 - 2x - 3 \geq 0\).
02

Factor the quadratic expression

We can factor the quadratic expression as follows: \((x-3)(x+1)\). So, the inequality becomes \((x-3)(x+1) \geq 0\).
03

Find the critical points

The critical points are the solutions to the equation \((x-3)(x+1) = 0\). We have two critical points here: \(x = 3\) and \(x = -1\).
04

Analyze the intervals between the critical points

To determine whether the inequality is true for an interval between the critical points, we can test the sign of the quadratic expression at any point within that interval: 1. For \(x < -1\), choose \(x = -2\): \((-2-3)(-2+1) = (-5)(-1) = 5 > 0\); the inequality is true in this interval. 2. For \(-1 < x < 3\), choose \(x = 0\): \((0-3)(0+1) = (-3)(1) = -3 < 0\); the inequality is false in this interval. 3. For \(x > 3\), choose \(x = 4\): \((4-3)(4+1) = (1)(5) = 5 > 0\); the inequality is true in this interval.
05

Write the final domain

Combining the intervals from Step 4, we can express the domain of the function as \[x \in (-\infty, -1] \cup [3, \infty)\]

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