Let `f: RrarrR` be defined as `f(x) = x^3+x-5`
If g(x) is a function such that `f(g(x))=x, AA 'x' in R` , then g'(63) is equal to ______

Options

A. `1/49`
B. `3/49`
C. `43/49`
D. `91/49`

Answer

To solve the problem, we need to find \( g'(63) \) given that \( f(g(x)) = x \) where \( f(x) = x^3 + x - 5 \).

### Step-by-Step Solution:

1. **Understanding the relationship**: We know that \( f(g(x)) = x \). This indicates that \( g(x) \) is the inverse function of \( f(x) \). Therefore, we can also express this relationship as \( g(f(x)) = x \).

2. **Differentiating both sides**: We differentiate the equation \( f(g(x)) = x \) with respect to \( x \):
\[ \frac{d}{dx}[f(g(x))] = \frac{d}{dx}[x] \]
Using the chain rule, we get:
\[ f'(g(x)) \cdot g'(x) = 1 \]

3. **Finding \( g'(x) \)**: Rearranging the equation gives us:
\[ g'(x) = \frac{1}{f'(g(x))} \]

4. **Finding \( g'(63) \)**: We need to find \( g'(63) \), so we substitute \( x = 63 \):
\[ g'(63) = \frac{1}{f'(g(63))} \]

5. **Finding \( g(63) \)**: To find \( g(63) \), we need to solve \( f(x) = 63 \):
\[ x^3 + x - 5 = 63 \]
Simplifying gives:
\[ x^3 + x - 68 = 0 \]

6. **Finding the root**: We can try \( x = 4 \):
\[ 4^3 + 4 - 68 = 64 + 4 - 68 = 0 \]
Thus, \( g(63) = 4 \).

7. **Calculating \( f'(x) \)**: Now we need to find \( f'(x) \):
\[ f'(x) = 3x^2 + 1 \]

8. **Finding \( f'(g(63)) \)**: Since \( g(63) = 4 \), we compute:
\[ f'(4) = 3(4^2) + 1 = 3(16) + 1 = 48 + 1 = 49 \]

9. **Final calculation of \( g'(63) \)**:
\[ g'(63) = \frac{1}{f'(g(63))} = \frac{1}{49} \]

Thus, the final answer is:
\[ \boxed{\frac{1}{49}} \]