If f(x) and g(x) are differentiable and increasing functions then which of the following functions alwasys increases?

Options

A. f(x)+g(x)
B. f(x)*g(x)
C. f(x)-g(x)
D. f(x)/g(x)

Explanation

To determine which of the given functions is always increasing when \( f(x) \) and \( g(x) \) are differentiable and increasing functions, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We know that \( f(x) \) and \( g(x) \) are differentiable and increasing functions. This means:
- \( f'(x) > 0 \)
- \( g'(x) > 0 \)

2. **Option A: \( h(x) = f(x) + g(x) \)**
- Differentiate \( h(x) \):
\[ h'(x) = f'(x) + g'(x) \]
- Since both \( f'(x) > 0 \) and \( g'(x) > 0 \), we have:
\[ h'(x) > 0 \]
- Therefore, \( h(x) \) is always increasing.

3. **Option B: \( h(x) = f(x) \cdot g(x) \)**
- Differentiate \( h(x) \):
\[ h'(x) = f'(x)g(x) + f(x)g'(x) \]
- Here, \( g(x) \) can be positive or negative since \( f(x) \) is not necessarily greater than zero. Thus, \( f'(x)g(x) \) can be positive or negative depending on the sign of \( g(x) \).
- Since we cannot guarantee that \( h'(x) > 0 \) for all \( x \), \( h(x) \) is not always increasing.

4. **Option C: \( h(x) = f(x) - g(x) \)**
- Differentiate \( h(x) \):
\[ h'(x) = f'(x) - g'(x) \]
- Both \( f'(x) \) and \( g'(x) \) are positive, but we cannot determine which is greater. Hence, \( h'(x) \) can be positive, negative, or zero depending on the relative magnitudes of \( f'(x) \) and \( g'(x) \).
- Therefore, \( h(x) \) is not guaranteed to be increasing.

5. **Option D: \( h(x) = \frac{f(x)}{g(x)} \)**
- Differentiate \( h(x) \):
\[ h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \]
- Here, \( g(x) \) can be positive or negative, and since \( f(x) \) can also be positive or negative, the numerator \( g(x)f'(x) - f(x)g'(x) \) can be positive, negative, or zero.
- Thus, \( h(x) \) is not guaranteed to be increasing.

### Conclusion:
The only function that is always increasing is:
- **Option A: \( h(x) = f(x) + g(x) \)**