Function `f(x), g(x)` are defined on `[-1, 3] and f''(x) > 0, g''(x) > 0` for all `x in [-1, 3]`, then which of the followingis always true ?

Options

A. f(x) -g(x) is concave upwards on (-1,3)
B. f(x) g(x) is concave upwards on (-1,3)
C. f(x) g(x) does not have a critical point on (-1,3)
D. f(x) +g (x) is concave upwards on (-1,3)

Answer

To solve the problem, we need to analyze the given functions \( f(x) \) and \( g(x) \) defined on the interval \([-1, 3]\) with the conditions that \( f''(x) > 0 \) and \( g''(x) > 0 \) for all \( x \) in that interval. This indicates that both functions are concave upward on the specified interval.

### Step-by-Step Solution:

1. **Understanding Concavity**:
- A function is concave upward if its second derivative is positive. Therefore, since \( f''(x) > 0 \) and \( g''(x) > 0 \), both functions \( f(x) \) and \( g(x) \) are concave upward.

2. **Analyzing the Options**:
- We need to evaluate the four options provided to determine which one is always true.

3. **Option 1: \( f(x) - g(x) \) is concave upward**:
- Let \( h(x) = f(x) - g(x) \).
- Then, \( h''(x) = f''(x) - g''(x) \).
- Since both \( f''(x) \) and \( g''(x) \) are positive, we cannot conclude that \( h''(x) > 0 \) because we do not know the relationship between \( f''(x) \) and \( g''(x) \). Hence, this option is not necessarily true.

4. **Option 2: \( f(x) \cdot g(x) \) is concave upward**:
- Let \( h(x) = f(x) \cdot g(x) \).
- Using the product rule, we have:
\[ h'(x) = f'(x)g(x) + f(x)g'(x) \]
- To find \( h''(x) \), we differentiate again:
\[ h''(x) = f''(x)g(x) + f'(x)g'(x) + f'(x)g'(x) + f(x)g''(x) \]
\[ h''(x) = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x) \]
- We cannot determine the sign of \( h''(x) \) without knowing the signs of \( f'(x) \) and \( g'(x) \). Thus, this option is also not necessarily true.

5. **Option 3: \( f(x) \cdot g(x) \) does not have critical points**:
- A critical point occurs when \( h'(x) = 0 \). Since we do not have information about \( f'(x) \) and \( g'(x) \), we cannot conclude that \( h'(x) \) is never zero. Therefore, this option is not necessarily true.

6. **Option 4: \( f(x) + g(x) \) is concave upward**:
- Let \( h(x) = f(x) + g(x) \).
- Then, \( h''(x) = f''(x) + g''(x) \).
- Since both \( f''(x) > 0 \) and \( g''(x) > 0 \), it follows that \( h''(x) > 0 \). Therefore, \( h(x) \) is concave upward.

### Conclusion:
The only statement that is always true is **Option 4: \( f(x) + g(x) \) is concave upward on \([-1, 3]\)**.