If `f'(x)=g(x)` and `g'(x)=-f(x)` for all `x` and `f(2)=4=f'(2)` then `f^(2)(4)+g^(2)(4)` is

Options

A. 8
B. 16
C. 32
D. 64

Explanation

To solve the problem, we start with the given conditions:

1. \( f'(x) = g(x) \)
2. \( g'(x) = -f(x) \)
3. \( f(2) = 4 \)
4. \( f'(2) = 4 \)

We need to find \( f^2(4) + g^2(4) \).

### Step 1: Differentiate \( f^2(x) + g^2(x) \)

We start by differentiating \( f^2(x) + g^2(x) \):

\[ \frac{d}{dx}(f^2(x) + g^2(x)) = \frac{d}{dx}(f^2(x)) + \frac{d}{dx}(g^2(x)) \]

Using the chain rule, we have:

\[ = 2f(x)f'(x) + 2g(x)g'(x) \]

### Step 2: Substitute \( f'(x) \) and \( g'(x) \)

Now we substitute \( f'(x) = g(x) \) and \( g'(x) = -f(x) \):

\[ = 2f(x)g(x) + 2g(x)(-f(x)) \]

This simplifies to:

\[ = 2f(x)g(x) - 2f(x)g(x) = 0 \]

### Step 3: Conclusion about \( f^2(x) + g^2(x) \)

Since the derivative \( \frac{d}{dx}(f^2(x) + g^2(x)) = 0 \), it implies that \( f^2(x) + g^2(x) \) is a constant function.

### Step 4: Find the constant value

To find the constant value, we can evaluate \( f^2(2) + g^2(2) \):

1. We know \( f(2) = 4 \).
2. Since \( f'(2) = g(2) = 4 \), we have \( g(2) = 4 \).

Now calculate:

\[ f^2(2) + g^2(2) = 4^2 + 4^2 = 16 + 16 = 32 \]

### Step 5: Use the constant value

Since \( f^2(x) + g^2(x) \) is constant, we have:

\[ f^2(4) + g^2(4) = f^2(2) + g^2(2) = 32 \]

### Final Answer

Thus, the value of \( f^2(4) + g^2(4) \) is:

\[ \boxed{32} \]

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