Let g be the inverse function of a differentiable function f and `G (x) =(1)/(g (x)).` If `f (4) =2` and `f '(4) =(1)/(16),` then the value of `(G'(2))^(2)` equals to:

Options

A. 1
B. 4
C. 16
D. 64

Answer

To solve the problem, we need to find the value of \( (G'(2))^2 \), where \( G(x) = \frac{1}{g(x)} \) and \( g \) is the inverse function of \( f \). We are given that \( f(4) = 2 \) and \( f'(4) = \frac{1}{16} \).

### Step-by-Step Solution:

1. **Understanding the relationship between \( f \) and \( g \)**:
Since \( g \) is the inverse of \( f \), we have:
\[ g(f(x)) = x \]
and
\[ f(g(x)) = x \]

2. **Differentiating the inverse function**:
By differentiating \( f(g(x)) = x \) with respect to \( x \), we apply the chain rule:
\[ f'(g(x)) \cdot g'(x) = 1 \]
This gives us:
\[ g'(x) = \frac{1}{f'(g(x))} \]

3. **Finding \( g(2) \)**:
We know from the problem statement that \( f(4) = 2 \). Since \( g \) is the inverse of \( f \), we have:
\[ g(2) = 4 \]

4. **Finding \( g'(2) \)**:
Now we can find \( g'(2) \) using the formula derived in step 2:
\[ g'(2) = \frac{1}{f'(g(2))} \]
Substituting \( g(2) = 4 \):
\[ g'(2) = \frac{1}{f'(4)} \]
Given that \( f'(4) = \frac{1}{16} \):
\[ g'(2) = \frac{1}{\frac{1}{16}} = 16 \]

5. **Calculating \( (G'(2))^2 \)**:
We know that \( G(x) = \frac{1}{g(x)} \). To find \( G'(x) \), we use the quotient rule:
\[ G'(x) = -\frac{g'(x)}{(g(x))^2} \]
Substituting \( x = 2 \):
\[ G'(2) = -\frac{g'(2)}{(g(2))^2} = -\frac{16}{(4)^2} = -\frac{16}{16} = -1 \]

6. **Final calculation**:
Now we find \( (G'(2))^2 \):
\[ (G'(2))^2 = (-1)^2 = 1 \]

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