1. Limits and Continuity

Continuity

1. Limits and Continuity

Continuity

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Multiple Choice

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

$open bracket negative 3 comma 0 close paren comma open bracket 0 comma infinity close paren$

$(-∞,0),[0,∞)$

$(-∞,∞)$

The function is not continuous anywhere.

Verified step by step guidance

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 $.

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 $.

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 $.

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 $.

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) $.

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