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June 28th, 2019, 09:08 PM   #1
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Constant function

Hello all,

What are the extreme values of the function $\displaystyle f(x)=C$ where $\displaystyle C$ is a real constant?

All the best,

Integrator

Last edited by Integrator; June 28th, 2019 at 09:19 PM.
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June 28th, 2019, 10:58 PM   #2
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The function's only value is $C$, which is the function's absolute maximum value and its absolute minimum value.
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June 29th, 2019, 09:29 PM   #3
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Quote:
Originally Posted by skipjack View Post
The function's only value is $C$, which is the function's absolute maximum value and its absolute minimum value.
Hello,

I'm sorry, but I do not understand!
It is true that from $\displaystyle f'(x)=0$ it follows that we have extremes, but $\displaystyle f''(x)=0$ does not show that there is any global minimum or maximum global...
If the function $\displaystyle f(x)=C$ has both the global minimum and the global maximum equal to $\displaystyle C$, then this means that the function $\displaystyle f(x)=C$ is both increasing and decreasing $\displaystyle \forall x\in \mathbb R$?!? Unbelievable and I do not know why mathematicians have reached this conclusion...
The function $\displaystyle f(x)=C$ has local extremes? I say it has no local extremes, and that's because $\displaystyle f''(x)=0$. Thank you very much!

All the best,

Integrator

Last edited by skipjack; July 1st, 2019 at 05:00 AM.
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June 30th, 2019, 08:25 AM   #4
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Quote:
Originally Posted by Integrator View Post
It is true that from $\displaystyle f'(x)=0$ it follows that we have extremes , but $\displaystyle f''(x)=0$ does not show that there is any global minimum or maximum global...
If the function $\displaystyle f(x)=C$ has both the global minimum and the global maximum equal to $\displaystyle C$ , then this means that the function $\displaystyle f(x)=C$ is both increasing and decreasing $\displaystyle \forall x\in \mathbb R$?!?[B]
You're confusing the simplified rules (almost always true) with the proper rules (more generally true, but worded more carefully).

$\displaystyle f'(x)= 0$ means the function is not changing locally (zero slope). This is a requirement for local minima/maxima, but not sufficient. (Look at $\displaystyle x^3\ @\ x=0$).

$\displaystyle f''(x)= 0$ means the slope is not changing locally (flat). If $\displaystyle f'(a) = 0\text{ and }f''(a) < 0$, then the function is increasing for $x<a$ and decreasing for $x>a$ (in the neighbourhood of a, anyway). This means it MUST be a local maximum. If $\displaystyle f'(a) = 0\text{ and }f''(a) > 0$, it MUST be a local minimum.

In contrast, $\displaystyle f'(x)=0\text{ and }f''(x)=0$ very often indicates neither a local minimum nor maximum, but this is not always the case. $\displaystyle f(x) = x^4\text{ and }f(x) = C$ both prove this.


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Originally Posted by Integrator View Post
If the function f(x)=C has both the global minimum and the global maximum equal to C , then this means that the function f(x)=C is both increasing and decreasing ∀x∈R?
I think both not decreasing and not increasing might be more accurate.

Would it help if we called $C$ the non-localised global minimum and the non-localised global maximum of $\displaystyle f(x)=C$?

Last edited by skipjack; July 1st, 2019 at 05:13 AM.
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June 30th, 2019, 08:45 PM   #5
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
You're confusing the simplified rules (almost always true) with the proper rules (more generally true, but worded more carefully).

$\displaystyle f'(x)= 0$ means the function is not changing locally (zero slope). This is a requirement for local minima/maxima, but not sufficient. (Look at $\displaystyle x^3\ @\ x=0$).

$\displaystyle f''(x)= 0$ means the slope is not changing locally (flat). If $\displaystyle f'(a) = 0\text{ and }f''(a) < 0$, then the function is increasing for $x<a$ and decreasing for $x>a$ (in the neighbourhood of a, anyway). This means it MUST be a local maximum. If $\displaystyle f'(a) = 0\text{ and }f''(a) > 0$, it MUST be a local minimum.

In contrast, $\displaystyle f'(x)=0\text{ and }f''(x)=0$ very often indicates neither a local minimum nor maximum, but this is not always the case. $\displaystyle f(x) = x^4\text{ and }f(x) = C$ both prove this.


Quote:
Originally Posted by Integrator View Post
If the function f(x)=C has both the global minimum and the global maximum equal to C , then this means that the function f(x)=C is both increasing and decreasing ∀x∈R?
I think both not decreasing and not increasing might be more accurate.

Would it help if we called $C$ the non-localised global minimum and the non-localised global maximum of $\displaystyle f(x)=C$?
Hello,

I think I do not confuse anything! In the case of $\displaystyle f(x)=x^4$, it is clear that the function has a global minimum because $\displaystyle f''(x)=12x^2>0$ $\displaystyle \forall x\neq 0$. In the case of $\displaystyle f(x)=C$, it follows that $\displaystyle f''(x)=0$ and that tells us nothing about the fact that the function $\displaystyle f(x)=C$ is increasing or is decreasing and so it seems incorrect to say that $\displaystyle f(x)=C$ is at the same time increasing and decreasing, which would mean that the function $\displaystyle f(x)=C$ would have the global minimum equal to the global maximum and equal with $\displaystyle C$....

All the best,

Integrator

Last edited by skipjack; July 1st, 2019 at 05:12 AM.
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June 30th, 2019, 09:09 PM   #6
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Math Focus: Algebraic Geometry
From an Analysis standpoint:

$\displaystyle f: \mathbb{R} \mapsto \{C\}$

max$\{C\}$ = min$\{C\} = C$
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