Math Problem Statement
Let f be the function defined as: f(x) = { a^2 + x^2 for x < 3, a(x + 3) for x ≥ 3 }, where a is a constant. For what values of a, if any, is f continuous at x = 3?
Solution
We are asked to determine for what values of , the given function is continuous at .
The function is defined as:
\begin{cases} a^2 + x^2 & \text{for } x < 3 \\ a(x + 3) & \text{for } x \geq 3 \end{cases}$$ To ensure continuity at $$x = 3$$, the following condition must hold: $$\lim_{x \to 3^-} f(x) = f(3) = \lim_{x \to 3^+} f(x)$$ ### Step 1: Evaluate $$f(3)$$ Since $$f(x) = a(x + 3)$$ for $$x \geq 3$$, we can directly substitute $$x = 3$$: $$f(3) = a(3 + 3) = a(6) = 6a$$ ### Step 2: Evaluate $$\lim_{x \to 3^-} f(x)$$ For $$x < 3$$, $$f(x) = a^2 + x^2$$. Taking the limit as $$x \to 3^-$$: $$\lim_{x \to 3^-} f(x) = a^2 + 3^2 = a^2 + 9$$ ### Step 3: Evaluate $$\lim_{x \to 3^+} f(x)$$ For $$x \geq 3$$, $$f(x) = a(x + 3)$$. Taking the limit as $$x \to 3^+$$: $$\lim_{x \to 3^+} f(x) = a(3 + 3) = 6a$$ ### Step 4: Set the limits equal for continuity For the function to be continuous at $$x = 3$$, we need: $$a^2 + 9 = 6a$$ ### Step 5: Solve for $$a$$ Rearranging the equation: $$a^2 - 6a + 9 = 0$$ This is a quadratic equation: $$(a - 3)^2 = 0$$ Thus, $$a = 3$$. ### Conclusion: The function $$f(x)$$ is continuous at $$x = 3$$ if and only if $$a = 3$$. Would you like more details or have any other questions? Here are 5 related questions to extend your understanding: 1. What are the conditions for a piecewise function to be continuous at a point? 2. How does the choice of $$a$$ affect the behavior of the function for $$x < 3$$? 3. Can a function be continuous but not differentiable at a point? If so, under what circumstances? 4. How do you handle continuity conditions for piecewise functions with more than two pieces? 5. What are the differences between left-hand and right-hand limits? **Tip:** When analyzing continuity, always check the left-hand and right-hand limits to ensure they match at the point in question!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Limits
Formulas
Continuity condition: lim (x -> c^-) f(x) = f(c) = lim (x -> c^+) f(x)
Quadratic formula for solving: ax^2 + bx + c = 0
Theorems
Continuity of a function at a point
Suitable Grade Level
Grades 10-12