Step 1: Understand the Domain of a Square Root Function The function \( f(x) = \sqrt{7 - 8x + x^2} \) is defined only when the expression inside the square root is non-negative. That is: \[ 7 - 8x + x^2 \geq 0. \] Step 2: Rewrite the Inequality Rewrite the inequality: \[ x^2 - 8x + 7 \geq 0. \] Step 3: Factor the Quadratic Expression Factor the quadratic expression: \[ x^2 - 8x + 7 = (x - 1)(x - 7). \] So, the inequality becomes: \[ (x - 1)(x - 7) \geq 0. \] Step 4: Solve the Inequality To solve \( (x - 1)(x - 7) \geq 0 \), we determine the intervals where the product is non-negative. The critical points are \( x = 1 \) and \( x = 7 \).
These divide the number line into three intervals:
1. \( x<1 \): Choose \( x = 0 \). Substituting into \( (x - 1)(x - 7) \): \[ (0 - 1)(0 - 7) = (-1)(-7) = 7>0. \] The product is positive in this interval.
2. \( 1<x<7 \): Choose \( x = 4 \). Substituting into \( (x - 1)(x - 7) \): \[ (4 - 1)(4 - 7) = (3)(-3) = -9<0. \] The product is negative in this interval.
3. \( x>7 \): Choose \( x = 8 \). Substituting into \( (x - 1)(x - 7) \): \[ (8 - 1)(8 - 7) = (7)(1) = 7>0. \] The product is positive in this interval.
Step 5: Include the Critical Points At \( x = 1 \) and \( x = 7 \), the expression equals zero, which satisfies the inequality \( \geq 0 \).
Therefore, the critical points are included in the solution.
Step 6: Write the Domain Combining the intervals where the product is non-negative, the domain of \( f(x) \) is: \[ (-\infty, 1] \cup [7, \infty). \]
Step 7: Verify the Answer The domain \( (-\infty, 1] \cup [7, \infty) \) corresponds to option (B).
Final Answer: The domain of \( f(x) \) is: \[ \boxed{(-\infty, 1] \cup [7, \infty)}. \]
Thus, the correct option is (B).
If $ X = A \times B $, $ A = \begin{bmatrix} 1 & 2 \\-1 & 1 \end{bmatrix} $, $ B = \begin{bmatrix} 3 & 6 \\5 & 7 \end{bmatrix} $, find $ x_1 + x_2 $.