Determine the interval(s) for which the function is continuous.

• A. $[-3,0),[0,\infty )$
• B. $(-∞,0),[0,∞)$
• C. $(-∞,∞)$
• D. The function is not continuous anywhere.

Answer: (A)

Verified step by step guidance

1. Step 1: Analyze the function piecewise. The function is defined as \( f(x) = \sqrt{9 - x^2} \) for \( -3 \leq x < 0 \), and \( f(x) = 5 \) for \( x \geq 0 \).
2. Step 2: Check the continuity of \( \sqrt{9 - x^2} \) on the interval \( -3 \leq x < 0 \). The square root function is continuous as long as the argument inside the square root is non-negative. Here, \( 9 - x^2 \geq 0 \) is satisfied for \( -3 \leq x \leq 3 \), so \( \sqrt{9 - x^2} \) is continuous on \( -3 \leq x < 0 \).
3. Step 3: Check the continuity of \( f(x) = 5 \) for \( x \geq 0 \). Since \( f(x) = 5 \) is a constant function, it is continuous for all \( x \geq 0 \).
4. Step 4: Examine the point \( x = 0 \), where the two pieces of the function meet. For \( x \to 0^- \), \( f(x) \to \sqrt{9 - 0^2} = 3 \). For \( x \to 0^+ \), \( f(x) = 5 \). Since the left-hand limit (3) does not equal the right-hand limit (5), the function is not continuous at \( x = 0 \).
5. Step 5: Combine the results. The function is continuous on \( [-3, 0) \) and \( [0, \infty) \), but not at \( x = 0 \). Therefore, the intervals of continuity are \( [-3, 0) \cup [0, \infty) \).