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Find the domain and the range of the fun...

Find the domain and the range of the function `y=f(x)`, where `f(x)` is given by
`sqrt(x^(2)-2x-3)`

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The correct Answer is:
To find the domain and range of the function \( y = f(x) = \sqrt{x^2 - 2x - 3} \), we will follow these steps: ### Step 1: Identify the expression under the square root The function is defined as: \[ f(x) = \sqrt{x^2 - 2x - 3} \] To ensure that \( f(x) \) is a real number, the expression inside the square root must be non-negative: \[ x^2 - 2x - 3 \geq 0 \] ### Step 2: Factor the quadratic expression We will factor the quadratic expression \( x^2 - 2x - 3 \): \[ x^2 - 2x - 3 = (x - 3)(x + 1) \] Thus, we need to solve the inequality: \[ (x - 3)(x + 1) \geq 0 \] ### Step 3: Determine the critical points The critical points occur where the expression equals zero: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] These points divide the number line into intervals. ### Step 4: Test the intervals We will test the sign of \( (x - 3)(x + 1) \) in the intervals determined by the critical points: 1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \): \[ (-2 - 3)(-2 + 1) = (-5)(-1) = 5 \quad \text{(positive)} \] 2. **Interval \( (-1, 3) \)**: Choose \( x = 0 \): \[ (0 - 3)(0 + 1) = (-3)(1) = -3 \quad \text{(negative)} \] 3. **Interval \( (3, \infty) \)**: Choose \( x = 4 \): \[ (4 - 3)(4 + 1) = (1)(5) = 5 \quad \text{(positive)} \] ### Step 5: Write the solution for the inequality The expression \( (x - 3)(x + 1) \geq 0 \) is satisfied in the intervals: \[ (-\infty, -1] \cup [3, \infty) \] Thus, the **domain** of \( f(x) \) is: \[ \text{Domain: } (-\infty, -1] \cup [3, \infty) \] ### Step 6: Determine the range of the function Since \( f(x) = \sqrt{x^2 - 2x - 3} \) is a square root function, it will yield non-negative values. The minimum value occurs at the endpoints of the domain: - At \( x = -1 \): \[ f(-1) = \sqrt{(-1)^2 - 2(-1) - 3} = \sqrt{1 + 2 - 3} = \sqrt{0} = 0 \] - At \( x = 3 \): \[ f(3) = \sqrt{3^2 - 2(3) - 3} = \sqrt{9 - 6 - 3} = \sqrt{0} = 0 \] As \( x \) approaches \( -\infty \) or \( +\infty \), \( f(x) \) approaches \( +\infty \). Thus, the **range** of \( f(x) \) is: \[ \text{Range: } [0, \infty) \] ### Final Answer - **Domain**: \( (-\infty, -1] \cup [3, \infty) \) - **Range**: \( [0, \infty) \)
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