If `f(x) = (2-(256 -7x)^(1//8))/((5x+32)^(1//5)-2), (x ne 2)`, then for f to be continuous everywhere, `f(0)` is equal to:

Options

A. `-1`
B. 1
C. `2^(6)`
D. `7//64`

Answer

To find the value of \( f(0) \) such that the function \( f(x) = \frac{2 - (256 - 7x)^{1/8}}{(5x + 32)^{1/5} - 2} \) is continuous everywhere, we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Set Up the Limit**:
We want to find \( f(0) \) such that:
\[ f(0) = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{2 - (256 - 7x)^{1/8}}{(5x + 32)^{1/5} - 2} \]

2. **Substituting \( x = 0 \)**:
Substitute \( x = 0 \) into the function:
\[ f(0) = \frac{2 - (256 - 7 \cdot 0)^{1/8}}{(5 \cdot 0 + 32)^{1/5} - 2} = \frac{2 - 256^{1/8}}{32^{1/5} - 2} \]
Calculate \( 256^{1/8} \) and \( 32^{1/5} \):
\[ 256^{1/8} = 2 \quad \text{(since \( 256 = 2^8 \))} \]
\[ 32^{1/5} = 2 \quad \text{(since \( 32 = 2^5 \))} \]
Thus,
\[ f(0) = \frac{2 - 2}{2 - 2} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

3. **Applying L'Hôpital's Rule**:
Differentiate the numerator and denominator:
- **Numerator**:
\[ \text{Let } g(x) = 2 - (256 - 7x)^{1/8} \]
\[ g'(x) = 0 - \frac{1}{8}(256 - 7x)^{-7/8} \cdot (-7) = \frac{7}{8}(256 - 7x)^{-7/8} \]
- **Denominator**:
\[ h(x) = (5x + 32)^{1/5} - 2 \]
\[ h'(x) = \frac{1}{5}(5x + 32)^{-4/5} \cdot 5 = (5x + 32)^{-4/5} \]

4. **Re-evaluate the Limit**:
Now we can evaluate the limit:
\[ \lim_{x \to 0} \frac{g'(x)}{h'(x)} = \lim_{x \to 0} \frac{\frac{7}{8}(256 - 7x)^{-7/8}}{(5x + 32)^{-4/5}} \]
Substituting \( x = 0 \):
\[ = \frac{\frac{7}{8}(256)^{-7/8}}{(32)^{-4/5}} = \frac{\frac{7}{8} \cdot \frac{1}{2^7}}{\frac{1}{2^4}} = \frac{\frac{7}{8} \cdot \frac{1}{128}}{\frac{1}{16}} = \frac{7}{8} \cdot \frac{16}{128} = \frac{7}{64} \]

5. **Conclusion**:
Therefore, for \( f \) to be continuous everywhere, we have:
\[ f(0) = \frac{7}{64} \]