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We are tasked with ensuring that the function \( f(x) \) is continuous at \( x = 2 \). The function is defined piecewise as:

\[
f(x) =
\begin{cases}
    c^2 x^2 & \text{if } x < 2 \\
    12cx + 28 & \text{if } x \geq 2
\end{cases}
\]

### Step 1: Continuity Condition
For the function to be continuous at \( x = 2 \), the left-hand limit and the right-hand limit must be equal to the function value at \( x = 2 \). This gives us the condition:
\[
\lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x)
\]

### Step 2: Left-hand Limit (\( x \to 2^- \))
Using the first part of the function, \( f(x) = c^2 x^2 \) for \( x < 2 \), the left-hand limit as \( x \to 2 \) is:
\[
\lim_{x \to 2^-} f(x) = c^2 (2)^2 = 4c^2
\]

### Step 3: Right-hand Limit (\( x \to 2^+ \))
For \( x \geq 2 \), \( f(x) = 12cx + 28 \). So, the right-hand limit as \( x \to 2 \) is:
\[
\lim_{x \to 2^+} f(x) = 12c(2) + 28 = 24c + 28
\]

### Step 4: Setting the Limits Equal
For continuity at \( x = 2 \), we set the left-hand limit equal to the right-hand limit:
\[
4c^2 = 24c + 28
\]

### Step 5: Solve the Equation
Now, we solve for \( c \). The equation is:
\[
4c^2 - 24c - 28 = 0
\]
This simplifies to:
\[
c^2 - 6c - 7 = 0
\]

We can solve this quadratic equation using the quadratic formula:
\[
c = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2(1)}
\]
\[
c = \frac{6 \pm \sqrt{36 + 28}}{2}
\]
\[
c = \frac{6 \pm \sqrt{64}}{2}
\]
\[
c = \frac{6 \pm 8}{2}
\]

Thus, we get two solutions:
\[
c = \frac{6 + 8}{2} = 7 \quad \text{or} \quad c = \frac{6 - 8}{2} = -1
\]

### Final Answer:
The values of \( c \) that make \( f(x) \) continuous at \( x = 2 \) are:
\[
c = 7 \quad \text{or} \quad c = -1
\]

Would you like more details or have any questions?

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Here are 5 related questions for further exploration:

1. What happens if we change the constant terms in the piecewise function?
2. Can we determine if \( f(x) \) is differentiable at \( x = 2 \)?
3. How would the continuity condition change if the function had more complex pieces?
4. What is the graphical interpretation of ensuring continuity for this function?
5. How does the quadratic formula help in solving continuity-related problems?

**Tip:** Always verify both left-hand and right-hand limits when checking for continuity in piecewise functions.

Consider the function f(x) = {c^2 * x^2 if x < 2; 12cx + 28 if x ≥ 2} where c is a constant. Find all value(s) of the constant c such that f(x) is continuous at x = 2.

Learn how to find the constant c that ensures the continuity of the piecewise function f(x) = {c^2 * x^2 if x < 2; 12cx + 28 if x ≥ 2} at x = 2. This problem involves limits, quadratic equations, and continuity conditions, ideal for Calculus I students.

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