About a monotonically decreasing f, does it exist a theorem that says

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SUMMARY

The discussion centers on the properties of the function y=2x*f(x), where f is a strictly monotonically decreasing, non-negative, derivable, and continuous function on the interval [0,c] with c≥1. The user seeks to establish whether a theorem exists that guarantees a single maximum point for the product of a monotonically increasing function (2x) and a monotonically decreasing function (f) under specific concavity conditions. The conclusion drawn is that if a continuous function has only one critical point, which is a local maximum, it must also be a global maximum due to the behavior of its derivative around that point.

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Homework Statement



I have y=2x*f(x).
f is strictly monotonically decreasing, non-negative, derivable and continuous in the close interval [0,c] with c>=1. it doesn't change its concavity in the interval, maybe beside at x=c/2. Note that 2x has the same properties but is monotonically increasing.
i've found a maximum point for y, at x=c/2, but not knowing f i was able only to demostrate that it was a local max. i want to say that it's the global max

2. The attempt at a solution

if there's a unique maximum for y in the interval [0,c], then my local max is automatically the global max

i think that is obviously true, but does it exist a theorem that says
the product of two nonnegative functions with fixed concavity, one monotonically increasing and the other decreasing in [a,b], is a function that has a single maximum point, being monotonically incresing before and monotonically decreasing after. if the max is in a or b then the new function is only increasing or decreasing
or something like that?
 
Last edited:
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Counterexample: f(x)=1/x. Or 1/sqrt(x). Or1/(x2)

It's true in general that if a function is continuous and has only a single critical point, which is a local max or min, then that is a global max or min. Why? Because the derivative to the left of that point must always be positive (it can only change sign at a critical point) and to the right it must be negative. Which means the function is increasing on [a,x] and decreasing on [x,b] where x is your critical point, and you can work out the details from there.
 
Thanks for the reply.

Office_Shredder said:
Counterexample: f(x)=1/x. Or 1/sqrt(x). Or1/(x2)

Sorry but i don't understand how these are counterexample. Beside that they aren't derivable in 0, if i multiply them with 2x i obtain functions that the max is in a or b and the new function is only increasing or decreasing

Probably i failed to write that 0=<a<b<+inf
 

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