If `f(x)` is continuous at `x=-2`, where `f(x)=(2)/(x+2)+(1)/(x^(2)-2x+4)-(24)/(x^(3)+8)`, for `x!= -2`, then `f(-2)=`

Options

A. `(-1)/(4)`
B. `1/4`
C. `(11)/(12)`
D. `(-11)/(12)`

Answer

To find the value of \( f(-2) \) for the function

\[ f(x) = \frac{2}{x+2} + \frac{1}{x^2 - 2x + 4} - \frac{24}{x^3 + 8} \]

given that \( f(x) \) is continuous at \( x = -2 \), we will follow these steps:

### Step 1: Identify the Form of the Function at \( x = -2 \)

First, we substitute \( x = -2 \) into the function to check the form:

\[ f(-2) = \frac{2}{-2 + 2} + \frac{1}{(-2)^2 - 2(-2) + 4} - \frac{24}{(-2)^3 + 8} \]

Calculating each term:

- The first term becomes \( \frac{2}{0} \) which is undefined.
- The second term becomes \( \frac{1}{4 - 4 + 4} = \frac{1}{4} \).
- The third term becomes \( \frac{24}{-8 + 8} = \frac{24}{0} \) which is also undefined.

Thus, both the first and third terms are undefined at \( x = -2 \), indicating that we need to simplify the function.

### Step 2: Simplify the Function

We need to find a limit as \( x \) approaches -2. We will combine the terms over a common denominator:

The common denominator for the first two terms is \( (x + 2)(x^2 - 2x + 4) \) and for the third term, it can be rewritten as \( (x + 2)(x^2 + 4) \).

So, we rewrite \( f(x) \):

\[ f(x) = \frac{2(x^2 - 2x + 4) + (x + 2) - 24}{(x + 2)(x^2 - 2x + 4)} \]

### Step 3: Substitute and Simplify

Now, we simplify the numerator:

1. Expand \( 2(x^2 - 2x + 4) = 2x^2 - 4x + 8 \).
2. The numerator becomes:

\[ 2x^2 - 4x + 8 + x + 2 - 24 = 2x^2 - 3x - 14 \]

So now we have:

\[ f(x) = \frac{2x^2 - 3x - 14}{(x + 2)(x^2 - 2x + 4)} \]

### Step 4: Factor the Numerator

Next, we need to factor \( 2x^2 - 3x - 14 \). We can use the quadratic formula or factorization. The factors are:

\[ 2x^2 - 3x - 14 = (2x + 7)(x - 2) \]

Thus, we can rewrite \( f(x) \):

\[ f(x) = \frac{(2x + 7)(x - 2)}{(x + 2)(x^2 - 2x + 4)} \]

### Step 5: Find the Limit as \( x \to -2 \)

Now we can find the limit as \( x \) approaches -2:

\[ \lim_{x \to -2} f(x) = \lim_{x \to -2} \frac{(2x + 7)(x - 2)}{(x + 2)(x^2 - 2x + 4)} \]

Substituting \( x = -2 \):

\[ = \frac{(2(-2) + 7)(-2 - 2)}{(-2 + 2)((-2)^2 - 2(-2) + 4)} \]

Calculating the numerator:

\[ = \frac{(-4 + 7)(-4)}{0 \cdot (4 + 4 + 4)} = \frac{3 \cdot (-4)}{0} = \text{undefined} \]

### Step 6: Apply L'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:

Differentiate the numerator and denominator:

1. Derivative of the numerator \( 2x^2 - 3x - 14 \) is \( 4x - 3 \).
2. Derivative of the denominator \( (x + 2)(x^2 - 2x + 4) \) requires the product rule:

\[ = (1)(x^2 - 2x + 4) + (x + 2)(2x - 2) \]

Now we can evaluate the limit again:

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

Calculating this limit gives us the value of \( f(-2) \).

### Final Result

After evaluating the limit, we find that:

\[ f(-2) = -\frac{11}{12} \]

Thus, the final answer is:

\[ \boxed{-\frac{11}{12}} \]