Determine whether the statement is true or false.
If [tex]$f$[/tex] is increasing and [tex]$f(x)>0$[/tex] on [tex]$I$[/tex], then [tex]$g(x)=\frac{1}{f(x)}$[/tex] is decreasing on [tex]$I$[/tex].
Answer (2)
Given that f is increasing and 0"> f ( x ) > 0 on interval I .
Consider x 1 < x 2 , which implies f ( x 1 ) < f ( x 2 ) because f is increasing.
Since f ( x 1 ) < f ( x 2 ) and both are positive, then \frac{1}{f(x_2)}"> f ( x 1 ) 1 > f ( x 2 ) 1 , meaning g(x_2)"> g ( x 1 ) > g ( x 2 ) .
Therefore, g ( x ) = f ( x ) 1 is decreasing on I , and the statement is True .
Explanation
Problem Analysis We are given that f is an increasing function and 0"> f ( x ) > 0 on the interval I . We want to determine if g ( x ) = f ( x ) 1 is decreasing on I . To do this, we will consider two arbitrary points x 1 and x 2 in the interval I such that x 1 < x 2 .
Using the Increasing Property of f(x) Since f is increasing on I , we know that if x 1 < x 2 , then f ( x 1 ) < f ( x 2 ) . Also, we are given that 0"> f ( x ) > 0 on I , so 0"> f ( x 1 ) > 0 and 0"> f ( x 2 ) > 0 .
Analyzing g(x) Now, let's consider g ( x 1 ) = f ( x 1 ) 1 and g ( x 2 ) = f ( x 2 ) 1 . Since f ( x 1 ) < f ( x 2 ) and both f ( x 1 ) and f ( x 2 ) are positive, we can take the reciprocal of both sides, which reverses the inequality: \frac{1}{f(x_2)}"> f ( x 1 ) 1 > f ( x 2 ) 1 . This means g(x_2)"> g ( x 1 ) > g ( x 2 ) .
Conclusion Since x 1 < x 2 implies g(x_2)"> g ( x 1 ) > g ( x 2 ) , we can conclude that g ( x ) is a decreasing function on the interval I . Therefore, the statement is true.
Examples
Imagine you're monitoring the population of a certain species of bacteria in a petri dish. If the bacteria population, represented by f ( x ) , is growing ( f is increasing) and always positive, then the amount of space each bacterium has, represented by g ( x ) = f ( x ) 1 , is decreasing. This is because as the number of bacteria increases, the space available per bacterium decreases, illustrating an inverse relationship.
The statement is true. If f is an increasing function and 0"> f ( x ) > 0 on interval I , then the function g ( x ) = f ( x ) 1 will be decreasing on that interval. This is because the reciprocal of an increasing positive function results in a decreasing function.
;
