How to find local maxima and minimas of undefined functions?

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To find local maxima and minima of the functions g(x) = |f(x)| and h(x) = +√f(x) over the interval −∞ < x < ∞, specific information about the function f(x) is essential. Without details on the properties of f(x), such as continuity and differentiability, it is impossible to determine local extrema. It is noted that g(x) is always non-negative, implying that any potential global minimum is at least zero. Additionally, for h(x) to be well-defined, f(x) must be non-negative for all x. Thus, further information about f(x) is necessary for a complete analysis.
jackgartlqn
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Hi,
I have been going over past papers and i found this question. Find any local maxima or local minima of the functions g(x) = |f(x)| and h(x) = +√f(x) in the interval, −∞ < x < ∞.
How Would I go about solving this?
 
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jackgartlqn said:
Hi,
I have been going over past papers and i found this question. Find any local maxima or local minima of the functions g(x) = |f(x)| and h(x) = +√f(x) in the interval, −∞ < x < ∞.
How Would I go about solving this?

Hey jackgartlqn and welcome to the forums.

You're going to need information about f(x). You will need some kind of information to start to answer this question in any kind of detail.

We do know that g(x) >= 0 no matter what which means in terms of a potential global minima it is always going to be >= 0 and if f(x) is >= 0 then we can say the same for h(x) as well.


Other than this though, you will need some kind of specific information for f(x) or at least for the properties of f(x).
 
jackgartlqn said:
Hi,
I have been going over past papers and i found this question. Find any local maxima or local minima of the functions g(x) = |f(x)| and h(x) = +√f(x) in the interval, −∞ < x < ∞.
How Would I go about solving this?



But for the fact that it MUST be f(x)\geq 0\,\,\forall x\in(-\infty,\infty) if h(x) is well defined, you can't deduce anything at all if you're not given

some more information about f(x) ...continuity, derivability,...?

DonAntonio
 
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