Question

Which of the following statements is always true?

• A. If $f(x)$ is decreasing, then $\frac{1}{f(x)}$ is increasing
• B. If $f(x)$ is decreasing, then $\frac{1}{f(x)}$ is also decreasing
• C. If both $f$ and $g$ are positive functions such that $f$ is decreasing and $g$ is increasing, then $\frac{f}{g}$ is a decreasing function
• D. If both $f$ and $g$ are positive functions such that $f$ is increasing and $g$ is decreasing, then $\frac{f}{g}$ is a decreasing function

Hint:

Think of it logically: if a fraction's numerator is shrinking (decreasing) while its denominator is simultaneously expanding (increasing), the overall value of the fraction must drop rapidly. Both actions work together to make the function decrease!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem tests the fundamental properties of monotonic functions (increasing and decreasing) and their behavior under division. A function \( h(x) \) is decreasing if its derivative \( h'(x)<0 \) for all \( x \) in its domain. When dealing with quotients of functions, we must apply the Quotient Rule of differentiation.
Step 2: Key Formula or Approach:
For a function \( h(x) = \frac{f(x)}{g(x)} \), the derivative is given by: \[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] We will analyze the sign of \( h'(x) \) based on the given conditions for \( f \) and \( g \).
Step 3: Detailed Explanation:
Let's analyze the statements one by one:
(A) and (B): Consider \( f(x) \). If \( f(x) \) is decreasing, then \( f'(x)<0 \). Let \( h(x) = \frac{1}{f(x)} \). Then \( h'(x) = -\frac{f'(x)}{(f(x))^2} \). The sign of \( h'(x) \) depends on the sign of \( f(x) \). If \( f(x) \) crosses the x-axis, the behavior of \( 1/f(x) \) changes abruptly near the root. For example, if \( f(x) = -x \), it is decreasing. \( 1/f(x) = -1/x \) is increasing for \( x>0 \) but the overall behavior is not strictly monotonic across the entire domain. Thus, (A) and (B) are not always true.
Now consider (C):
Given that \( f(x)>0 \), \( g(x)>0 \), \( f \) is decreasing (\( f'(x)<0 \)), and \( g \) is increasing (\( g'(x)>0 \)).
Let \( h(x) = \frac{f(x)}{g(x)} \). Its derivative is: \[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \] In the numerator: - Since \( f'(x)<0 \) and \( g(x)>0 \), the term \( f'(x)g(x) \) is negative.
- Since \( f(x)>0 \) and \( g'(x)>0 \), the term \( f(x)g'(x) \) is positive.
- Therefore, the subtraction \( (\text{negative term}) - (\text{positive term}) \) results in a negative value.
Thus, the numerator is always negative (\(<0 \)).
Since the denominator \( (g(x))^2 \) is always positive, the total derivative \( h'(x) \) is always negative.
Conclusion: \( \frac{f}{g} \) is a strictly decreasing function. This statement is always true.
Step 4: Final Answer:
By derivative analysis, we have proven that the ratio of a positive decreasing function to a positive increasing function is always decreasing. Therefore, option (C) is correct.