Can We Find an Elegant Solution For a Less Than Function?

[matthewknight]

A challenge:

For a function f(x), does there exist a function g(f(x)) such that the outputted function is always less than f(x)?

There are a couple of approaches, but the immediate ones are at first glance quite inelegant requiring stipulations to be introduced. The challenge is to find the most natural solution--the most elegant. Any takers?

Admittedly, there is probably an extremely simple answer that I am overlooking.

 

[mmmboh]

g(x) = -2x if x>0, 2x if x<0, -1 if x=0.

 

[matthewknight]

g(x) = -2x if x>0, 2x if x<0, -1 if x=0.

Those pesky, inelegant stipulations I was talking about... :)

 

[Mark44]

A challenge:

For a function f(x), does there exist a function g(f(x)) such that the outputted function is always less than f(x)?

What comes out is a number, not a function.

If g(x) = x - 1, g(f(x)) will always be less (by 1) than f(x), since g(f(x)) = f(x) - 1.

There are a couple of approaches, but the immediate ones are at first glance quite inelegant requiring stipulations to be introduced. The challenge is to find the most natural solution--the most elegant. Any takers?

Admittedly, there is probably an extremely simple answer that I am overlooking.

 

[raam86]

After analyzing the function and knowing their maximum\minimum etc' you can show that
when g(x)>/</=f(x)

This is how you approach this in high school anyway...

 

[Mark44]

After analyzing the function and knowing their maximum\minimum etc' you can show that
when g(x)>/</=f(x)

What does this notation (>/</=) mean?

Is this saying that g(x) > f(x) OR g(x) < f(x) OR g(x) = f(x)?
If so, that doesn't tell us much. The Archimedean Trichotomy says that given any two real numbers a and b, then exactly one of the following must be true.
1. a < b
2. a = b
3. a > b

 

[raam86]

Of course this is true. What I mean is ,assuming this is high school - college level etc', After you know max\min and "up\down domain"*
you will know which function is "over" or under".
let y=x+2 be f(x) and y=ln(x) g(x)
http://www.wolframalpha.com/input/?i=plot+y=ln(x),+y=x+2&asynchronous=false&equal=Submit

It is easy to so that when x>-1.5 f(x)>g(x)

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When doing it by hand you can find the derivatives, x\y 0 coordinates etc'*

*Sorry got a language problem here. Not sure what is the technical term in English.