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To ensure the function \( f(x) \) is continuous, we need to guarantee that the function values match at the point where \( x = 1 \), since the function is defined differently for \( x < 1 \) and \( x \geq 1 \).

### Step-by-step solution:

1. **Left-hand limit** \( x \to 1^- \):
   For \( x < 1 \), \( f(x) = 3 - x^2 \).
   \[
   \lim_{x \to 1^-} f(x) = 3 - (1)^2 = 3 - 1 = 2
   \]

2. **Right-hand limit** \( x \to 1^+ \):
   For \( x \geq 1 \), \( f(x) = ax^2 - 1 \).
   \[
   \lim_{x \to 1^+} f(x) = a(1)^2 - 1 = a - 1
   \]

3. **Continuity condition**:
   For continuity at \( x = 1 \), the left-hand limit must equal the right-hand limit:
   \[
   2 = a - 1
   \]

4. **Solve for \( a \)**:
   \[
   a = 2 + 1 = 3
   \]

Thus, the value of \( a \) that makes the function \( f(x) \) continuous is \( \boxed{3} \).

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Do you want further details or have any questions? Here are five related problems to expand your understanding:

1. What happens if we try a different value of \( a \)? Is there any discontinuity?
2. How would you check the continuity of a piecewise function if there were three pieces?
3. What are the conditions for differentiability at \( x = 1 \) for the same function?
4. How can you generalize this problem to other points of discontinuity?
5. How would you extend this function to ensure it's smooth (differentiable) everywhere?

**Tip**: To ensure a function is continuous, always verify that the left-hand and right-hand limits are equal at the point of interest.

Find a value for a so that the function f(x) = {3 - x^2, x < 1; ax^2 - 1, x ≥ 1} is continuous.

Learn how to find a value for 'a' to ensure the continuity of a piecewise function. This problem involves limits and continuity, suitable for high school students studying advanced algebra or calculus.

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