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Let g'(x)gt 0 and f'(x) lt 0 AA x in R, ...

Let `g'(x)gt 0 and f'(x) lt 0 AA x in R`, then

A

`g(f(x+1))gt g(f(x-1))`

B

`f(g(x-1)) lt f(g(x+1))`

C

`g(f(x+1)) lt g(f(x-1))`

D

`g(g'(x+1)) lt g(g(x+1))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the implications of the given conditions on the functions \( g(x) \) and \( f(x) \). ### Step-by-Step Solution: 1. **Understanding the Derivatives**: - We are given that \( g'(x) > 0 \) for all \( x \in \mathbb{R} \). This means that \( g(x) \) is a strictly increasing function. - We are also given that \( f'(x) < 0 \) for all \( x \in \mathbb{R} \). This means that \( f(x) \) is a strictly decreasing function. **Hint**: Recall that if a function's derivative is positive, the function is increasing; if it's negative, the function is decreasing. 2. **Analyzing the Expression**: - We need to analyze the expression \( g(f(x) + 1) \) and \( g(f(x) - 1) \). - Since \( f(x) \) is strictly decreasing, if \( x_1 < x_2 \), then \( f(x_1) > f(x_2) \). Therefore, \( f(x) + 1 \) and \( f(x) - 1 \) will also respect this order. **Hint**: Consider how the properties of increasing and decreasing functions affect the order of their outputs. 3. **Applying the Increasing Property of \( g \)**: - Since \( g \) is strictly increasing, we have: \[ f(x) + 1 > f(x) - 1 \implies g(f(x) + 1) > g(f(x) - 1) \] - This means that \( g(f(x) + 1) > g(f(x) - 1) \). **Hint**: Use the property of increasing functions to compare the outputs based on their inputs. 4. **Conclusion**: - The statement \( g(f(x) + 1) > g(f(x) - 1) \) is true based on the properties of \( g \) and \( f \). - Therefore, the conclusion we can draw from the given information is that the statement holds true. ### Final Answer: The statement \( g(f(x) + 1) > g(f(x) - 1) \) is true.
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