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

Options

A. `-1`
B. 1
C. `2^6`
D. none of these

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 on the interval \([-1, 1]\), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Substituting \( x = 0 \)**:
\[ f(0) = \frac{2 - (256 - 7 \cdot 0)^{1/8}}{(5 \cdot 0 + 32)^{1/5} - 2} = \frac{2 - (256)^{1/8}}{(32)^{1/5} - 2} \]

2. **Calculating \( (256)^{1/8} \)**:
\[ (256)^{1/8} = (2^8)^{1/8} = 2 \]
Thus, the numerator becomes:
\[ 2 - 2 = 0 \]

3. **Calculating \( (32)^{1/5} \)**:
\[ (32)^{1/5} = (2^5)^{1/5} = 2 \]
Thus, the denominator becomes:
\[ 2 - 2 = 0 \]

4. **Indeterminate Form**:
Since both the numerator and denominator approach 0, we have the indeterminate form \( \frac{0}{0} \). We will apply L'Hôpital's Rule.

5. **Applying L'Hôpital's Rule**:
We differentiate the numerator and denominator separately.
- **Numerator**:
\[ \text{Let } g(x) = 2 - (256 - 7x)^{1/8} \]
The derivative \( g'(x) \) is:
\[ g'(x) = 0 - \frac{1}{8}(256 - 7x)^{-7/8} \cdot (-7) = \frac{7}{8}(256 - 7x)^{-7/8} \]

- **Denominator**:
\[ \text{Let } h(x) = (5x + 32)^{1/5} - 2 \]
The derivative \( h'(x) \) is:
\[ h'(x) = \frac{1}{5}(5x + 32)^{-4/5} \cdot 5 = (5x + 32)^{-4/5} \]

6. **Taking the Limit Again**:
Now we compute 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 \( x = 0 \):
\[ = \frac{\frac{7}{8}(256)^{-7/8}}{(32)^{-4/5}} = \frac{\frac{7}{8}(2)^{-7}}{(2)^{-4}} = \frac{\frac{7}{8} \cdot \frac{1}{128}}{\frac{1}{16}} = \frac{7}{8} \cdot \frac{16}{128} = \frac{7}{8} \cdot \frac{1}{8} = \frac{7}{64} \]

### Conclusion:
Thus, for \( f \) to be continuous at \( x = 0 \), we have:
\[ f(0) = \frac{7}{64} \]