Value of `f(-2),` if f(x) = ` (x^(3)+8)/(x^(5)+32)` is continuous at x = `-2` is

Options

A. `-3/20`
B. `3/20`
C. `20/3`
D. `-20/3`

Answer

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

\[ f(x) = \frac{x^3 + 8}{x^5 + 32} \]

given that it is continuous at \( x = -2 \), we need to ensure that

\[ \lim_{x \to -2} f(x) = f(-2). \]

### Step 1: Evaluate \( f(-2) \)

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

\[ f(-2) = \frac{(-2)^3 + 8}{(-2)^5 + 32} = \frac{-8 + 8}{-32 + 32} = \frac{0}{0}. \]

Since this results in an indeterminate form \( \frac{0}{0} \), we need to apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator:

1. **Numerator**: The derivative of \( x^3 + 8 \) is \( 3x^2 \).
2. **Denominator**: The derivative of \( x^5 + 32 \) is \( 5x^4 \).

So we rewrite the limit:

\[ \lim_{x \to -2} f(x) = \lim_{x \to -2} \frac{3x^2}{5x^4}. \]

### Step 3: Substitute \( x = -2 \) Again

Now we substitute \( x = -2 \) into the derivatives:

\[ \lim_{x \to -2} \frac{3x^2}{5x^4} = \frac{3(-2)^2}{5(-2)^4} = \frac{3 \cdot 4}{5 \cdot 16} = \frac{12}{80} = \frac{3}{20}. \]

### Conclusion

Thus, for the function to be continuous at \( x = -2 \), we have:

\[ f(-2) = \frac{3}{20}. \]

### Final Answer

The value of \( f(-2) \) is

\[ \frac{3}{20}. \]