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To solve this problem, we need to ensure the function \( f(x) \) is continuous at \( x = 2 \). Continuity at \( x = 2 \) requires that the left-hand limit and right-hand limit of \( f(x) \) as \( x \) approaches 2 are equal, and that they also equal the function's value at \( x = 2 \).

Given the piecewise function:
\[
f(x) = \begin{cases} 
cx^2 + 3x & \text{if } x < 2 \\
x^3 - cx & \text{if } x \geq 2 
\end{cases}
\]

### Step 1: Find the Left-Hand Limit as \( x \to 2^- \)
The left-hand limit as \( x \) approaches 2 from the left is:
\[
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (cx^2 + 3x)
\]
Substitute \( x = 2 \) into the equation:
\[
\lim_{x \to 2^-} f(x) = c(2)^2 + 3(2) = 4c + 6
\]

### Step 2: Find the Right-Hand Limit as \( x \to 2^+ \)
The right-hand limit as \( x \) approaches 2 from the right is:
\[
\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^3 - cx)
\]
Substitute \( x = 2 \) into the equation:
\[
\lim_{x \to 2^+} f(x) = (2)^3 - c(2) = 8 - 2c
\]

### Step 3: Set the Limits Equal for Continuity
For the function to be continuous at \( x = 2 \), the left-hand limit and right-hand limit must be equal:
\[
4c + 6 = 8 - 2c
\]
Now, solve for \( c \):
\[
4c + 2c = 8 - 6
\]
\[
6c = 2
\]
\[
c = \frac{1}{3}
\]

So, the value of the constant \( c \) that makes the function continuous on \( (-\infty, \infty) \) is \( c = \frac{1}{3} \).

If you have any further questions or need more details, feel free to ask!

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### Related Questions:
1. How does the value of \( c \) affect the shape of the graph for \( x < 2 \) and \( x \geq 2 \)?
2. What happens to the function if \( c \) is not equal to \( \frac{1}{3} \)?
3. How do you check for differentiability at \( x = 2 \)?
4. Can the function be discontinuous at any other point besides \( x = 2 \)?
5. How would you find the derivative of the function for both intervals?

### Tip:
Always remember to check both the left-hand and right-hand limits when verifying continuity at a particular point in a piecewise function.

For what value of the constant c is the function f continuous on (-∞, ∞)? The function f(x) is given by: f(x) = cx^2 + 3x if x < 2, and f(x) = x^3 - cx if x ≥ 2. The function is continuous on (-∞, 2) and (2, ∞). To ensure continuity on (-∞, ∞), we need to check the limits as x approaches 2 from both sides.

This problem requires finding the value of the constant c that ensures the function f(x) is continuous over the entire real line. The function is piecewise, with different expressions for x < 2 and x ≥ 2. Learn how to use limits to determine the value of c and ensure continuity at x = 2.

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