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The problem asks to find the value of the constant \( a \) that makes the function \( f(x) \) continuous on \( (-\infty, \infty) \). The function is piecewise defined as:

\[
f(x) = \begin{cases} 
\frac{4x^3 + 4x^2 - 8x}{x + 2} & \text{if } x < -2 \\
-3x^2 - 5x + a & \text{if } x \geq -2 
\end{cases}
\]

### Step 1: Continuity Condition
For the function to be continuous at \( x = -2 \), the left-hand limit as \( x \to -2^- \) must equal the right-hand limit as \( x \to -2^+ \), and they must both equal \( f(-2) \).

Thus, we need:
\[
\lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x)
\]

### Step 2: Left-hand Limit (as \( x \to -2^- \))

We consider the expression for \( f(x) \) when \( x < -2 \), which is:
\[
f(x) = \frac{4x^3 + 4x^2 - 8x}{x + 2}
\]
At \( x = -2 \), the denominator \( x + 2 \) becomes zero, so this expression is indeterminate. Let's simplify it by factoring the numerator:
\[
4x^3 + 4x^2 - 8x = 4x(x^2 + x - 2) = 4x(x - 1)(x + 2)
\]
Thus, we have:
\[
f(x) = \frac{4x(x - 1)(x + 2)}{x + 2}
\]
Canceling \( (x + 2) \) (for \( x \neq -2 \)):
\[
f(x) = 4x(x - 1)
\]
Now, take the limit as \( x \to -2^- \):
\[
\lim_{x \to -2^-} f(x) = 4(-2)((-2) - 1) = 4(-2)(-3) = 24
\]

### Step 3: Right-hand Limit (as \( x \to -2^+ \))

For \( x \geq -2 \), we use the second expression for \( f(x) \):
\[
f(x) = -3x^2 - 5x + a
\]
Substitute \( x = -2 \):
\[
f(-2) = -3(-2)^2 - 5(-2) + a = -3(4) + 10 + a = -12 + 10 + a = -2 + a
\]

### Step 4: Continuity Condition at \( x = -2 \)

For continuity, the left-hand limit must equal the right-hand limit:
\[
24 = -2 + a
\]
Solving for \( a \):
\[
a = 24 + 2 = 26
\]

Thus, the value of \( a \) is \( \boxed{26} \).

Would you like more details or have any questions? Here are five related questions to deepen your understanding:

1. How do you determine if a function is continuous at a point?
2. Why was it necessary to factor the numerator in the left-hand limit?
3. What would happen if the limits from the left and right did not match?
4. Can you apply the same technique to find discontinuities in other types of piecewise functions?
5. What other conditions, apart from continuity, could be checked at boundary points in piecewise functions?

**Tip:** When dealing with piecewise functions, always verify the conditions for continuity, differentiability, and any other constraints at the boundaries of the defined intervals.

Find the value of the constant a that makes the following function continuous on (-∞, ∞):

f(x) = { (4x^3 + 4x^2 - 8x)/(x + 2), if x < -2
          -3x^2 - 5x + a, if x ≥ -2 }


Learn how to find the value of the constant 'a' that makes the function f(x) continuous on (-∞, ∞). This step-by-step solution covers limits, factoring polynomials, and applying the continuity condition at boundary points. Ideal for pre-calculus or calculus students.

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