Question

The domain of the function \(f(x) = (x^2-2x-63)^{\frac{3}{2}}\), x ∈ R is

• A. (-∞, -6]∪[9,∞)
• B. (-∞, -9]∪[7,∞)
• C. (-∞, -7]∪[7,∞)
• D. (-∞, -5]∪[9,∞)
• E. (-∞, -7]∪[9,∞)

Answer 1

Answer: (E)

For the function \( f(x) = \left( x^2 - 2x - 63 \right)^{3/2} \), the domain requires the expression inside the square root, \( x^2 - 2x - 63 \), to be non-negative. Thus, we need to solve the inequality: \[ x^2 - 2x - 63 \geq 0 \] Factoring the quadratic expression: \[ (x - 9)(x + 7) \geq 0 \] The solution to this inequality is \( x \in (-\infty, -7] \cup [9, \infty) \)

The correct option is (E) : \((-∞, -7]∪[9,∞)\)

Answer 2

We have the function \(f(x) = (x^2 - 2x - 63)^{\frac{3}{2}}\). Since we're taking a real number to the power of 3/2, the base \(x^2 - 2x - 63\) must be non-negative to avoid complex numbers.

Therefore, we need to solve the inequality: \(x^2 - 2x - 63 \geq 0\)

First, we factor the quadratic expression:

\(x^2 - 2x - 63 = (x - 9)(x + 7)\)

Now we need to solve \((x - 9)(x + 7) \geq 0\). The roots of the quadratic are x = 9 and x = -7. We can use a sign chart to determine when the expression is non-negative:

Interval	x + 7	x - 9	(x + 7)(x - 9)
x < -7	-	-	+
-7 < x < 9	+	-	-
x > 9	+	+	+

From the sign chart, we can see that the expression is non-negative when x ≤ -7 or x ≥ 9.

Thus, the domain of the function is (-∞, -7] ∪ [9, ∞).

Therefore, the answer is (-∞, -7] ∪ [9, ∞).