Let `f: [1,3] to R` be a continuous function that is differentiable in `(1,3) and f '(x) = |f(x) |^(2) + 4` for all `x in (1,3).` Then,

Options

A. `f (3) -f (1) =5` is true
B. `f (3) -f (1) =5` is false
C. `f (3)-f (1) =7` is false
D. `f (3) -f (1) lt 0`only at one point of `(1,3)`

Answer

To solve the problem, we will apply Lagrange's Mean Value Theorem (LMVT) and analyze the given derivative condition.

### Step-by-Step Solution:

1. **Understanding the Function and Its Derivative**:
We are given a function \( f: [1, 3] \to \mathbb{R} \) that is continuous on the closed interval \([1, 3]\) and differentiable on the open interval \( (1, 3) \). The derivative is given by:
\[ f'(x) = |f(x)|^2 + 4 \]
for all \( x \in (1, 3) \).

2. **Applying Lagrange's Mean Value Theorem**:
According to the LMVT, there exists at least one point \( c \in (1, 3) \) such that:
\[ f'(c) = \frac{f(3) - f(1)}{3 - 1} = \frac{f(3) - f(1)}{2} \]

3. **Substituting the Derivative**:
From the derivative condition, we know:
\[ f'(c) = |f(c)|^2 + 4 \]
Therefore, we can equate the two expressions:
\[ |f(c)|^2 + 4 = \frac{f(3) - f(1)}{2} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ f(3) - f(1) = 2(|f(c)|^2 + 4) \]
Simplifying this, we have:
\[ f(3) - f(1) = 2|f(c)|^2 + 8 \]

5. **Analyzing the Expression**:
Since \( |f(c)|^2 \geq 0 \) (as it is a square), the minimum value of \( 2|f(c)|^2 \) is \( 0 \). Thus:
\[ f(3) - f(1) \geq 8 \]

6. **Conclusion**:
Therefore, we conclude that:
\[ f(3) - f(1) \geq 8 \]
This means that the difference between the values of the function at 3 and 1 is at least 8.

### Final Answer:
The correct conclusion is that \( f(3) - f(1) \) is always greater than or equal to 8.