Let f and g be two differentiable functions defined on an interval I such that `f(x)>=0` and `g(x)<= 0` for all `x in I` and f is strictly decreasing on I while g is strictly increasing on I then (A) the product function fg is strictly increasing on I (B) the product function fg is strictly decreasing on I (C) fog(x) is monotonically increasing on I (D) fog (x) is monotonically decreasing on I

Options

A. The product function fg is strictly increasing on I
B. The product function fg is strictly decreasing on I
C. fog (x) is monotonically increasing on I
D. fog (x) is monotonically decreasing onI

Answer

To solve the problem, we need to analyze the behavior of the functions \( f \) and \( g \) based on the given conditions. Let's break down the solution step by step.

### Step 1: Understand the properties of \( f \) and \( g \)
- We know that \( f(x) \geq 0 \) for all \( x \in I \) and \( f \) is strictly decreasing. This means that \( f'(x) < 0 \) for all \( x \in I \).
- We also know that \( g(x) \leq 0 \) for all \( x \in I \) and \( g \) is strictly increasing. This means that \( g'(x) > 0 \) for all \( x \in I \).

### Step 2: Analyze the product \( fg \)
We need to determine if the product \( fg \) is strictly increasing or strictly decreasing on the interval \( I \).

Using the product rule for differentiation:
\[ (fg)'(x) = f'(x)g(x) + f(x)g'(x) \]

### Step 3: Evaluate the terms in the derivative
- Since \( f'(x) < 0 \) and \( g(x) \leq 0 \), the term \( f'(x)g(x) \) will be non-positive (because a negative times a non-positive is non-positive).
- Since \( f(x) \geq 0 \) and \( g'(x) > 0 \), the term \( f(x)g'(x) \) will be non-negative (because a non-negative times a positive is non-negative).

### Step 4: Combine the terms
Now we have:
\[ (fg)'(x) = f'(x)g(x) + f(x)g'(x) \]
- The first term \( f'(x)g(x) \) is non-positive.
- The second term \( f(x)g'(x) \) is non-negative.

Since \( f'(x)g(x) \) is negative or zero and \( f(x)g'(x) \) is positive or zero, the sum \( (fg)'(x) \) will be negative or zero, indicating that \( fg \) is strictly decreasing on \( I \).

### Conclusion for options A and B
- Therefore, option (A) is incorrect and option (B) is correct: the product function \( fg \) is strictly decreasing on \( I \).

### Step 5: Analyze the composition \( f(g(x)) \)
Next, we check the options regarding the composition \( f(g(x)) \).

Using the chain rule:
\[ \frac{d}{dx} f(g(x)) = f'(g(x))g'(x) \]
- Since \( g'(x) > 0 \) (increasing function) and \( f'(g(x)) < 0 \) (decreasing function), the product \( f'(g(x))g'(x) \) will be negative.

### Conclusion for options C and D
- This means that \( f(g(x)) \) is strictly decreasing on \( I \), making option (D) correct and option (C) incorrect.

### Final Answers
- (A) Incorrect
- (B) Correct
- (C) Incorrect
- (D) Correct