At what points is the following function​ continuous?

​f(x)equals=left brace Start 3 By 2 Matrix 1st Row 1st Column StartFraction x cubed plus 8 Over x squared minus 4 EndFraction comma 2nd Column x not equals 2 comma x not equals minus 2 2nd Row 1st Column negative 3 comma 2nd Column x equals negative 2 3rd Row 1st Column 8 comma 2nd Column x equals 2 EndMatrix

x3+8x2−4,

x≠2, x≠−2

−3,

x=−2

8,

x=2

Question content area bottom

Part 1

Choose the correct answer below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

All real numbers except at 

xequals=enter your response here

​(Use a comma to separate answers as​ needed.)

B.

Only at 

xequals=enter your response here

​(Use a comma to separate answers as​ needed.)

C.

All real numbers x

We are given a piecewise function, and we need to determine at which points this function is continuous. Let’s break it down.

The function \( f(x) \) is defined as follows:

\[
f(x) =
\begin{cases}
\frac{x^3 + 8}{x^2 - 4}, & \text{if } x \neq 2 \text{ and } x \neq -2 \\
-3, & \text{if } x = -2 \\
8, & \text{if } x = 2
\end{cases}
\]

### Step 1: Analyze \( f(x) \) when \( x \neq 2 \) and \( x \neq -2 \)

The function is given by:

\[
f(x) = \frac{x^3 + 8}{x^2 - 4} = \frac{x^3 + 8}{(x-2)(x+2)}
\]

This rational function has a potential issue at the points where the denominator is zero, which happens when \( x = 2 \) or \( x = -2 \).

Thus, \( f(x) \) has **discontinuities at \( x = 2 \) and \( x = -2 \)** for now.

### Step 2: Check the behavior at \( x = -2 \)

At \( x = -2 \), the function value is explicitly given as \( f(-2) = -3 \). To check for continuity, we need the limit as \( x \) approaches \( -2 \) from both sides. First, simplify \( f(x) \):

\[
f(x) = \frac{(x + 2)(x^2 - 2x + 4)}{(x-2)(x+2)} = \frac{x^2 - 2x + 4}{x - 2} \text{ for } x \neq 2 \text{ and } x \neq -2
\]

Now, evaluate the limit as \( x \to -2 \):

\[
\lim_{x \to -2} f(x) = \lim_{x \to -2} \frac{x^2 - 2x + 4}{x - 2}
\]

Substitute \( x = -2 \):

\[
\lim_{x \to -2} f(x) = \frac{(-2)^2 - 2(-2) + 4}{-2 - 2} = \frac{4 + 4 + 4}{-4} = -3
\]

Since \( f(-2) = -3 \), the function is continuous at \( x = -2 \).

### Step 3: Check the behavior at \( x = 2 \)

At \( x = 2 \), the function value is explicitly given as \( f(2) = 8 \). Now, we check the limit as \( x \to 2 \):

\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^2 - 2x + 4}{x - 2}
\]

Substituting \( x = 2 \) gives an undefined form, but we can evaluate the limit algebraically.

Notice that the function cannot be simplified further to avoid the discontinuity at \( x = 2 \), meaning \( \lim_{x \to 2} f(x) \) does not exist in a way that equals \( 8 \). Therefore, **the function is discontinuous at \( x = 2 \)**.

### Conclusion:

- \( f(x) \) is continuous at all points except \( x = 2 \) and \( x = -2 \).

Thus, the correct answer is:

\[
\boxed{A. \text{All real numbers except at } x = 2, -2}
\]

---

Would you like further details or have any questions?

### Related questions:
1. How do you prove that a function is continuous at a given point?
2. What are the conditions for a rational function to be continuous?
3. How can we use limits to determine discontinuity?
4. What happens at points where the denominator of a function is zero?
5. How do piecewise functions behave at the boundaries of their definitions?

### Tip:
When dealing with rational functions, always check for points where the denominator is zero, as these are potential discontinuities.

At what points is the following function continuous? f(x) = { (x^3 + 8)/(x^2 - 4), x ≠ 2, x ≠ -2; -3, x = -2; 8, x = 2 }.

This problem involves finding the points of continuity for a piecewise rational function. It focuses on discontinuities at x = 2 and x = -2, using concepts such as limits, rational functions, and the Continuity Theorem. A step-by-step solution is provided, suitable for students in Grades 11-12.

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