Constant function

Aug 2018
137
7
România
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:

skipjack

Forum Staff
Dec 2006
21,479
2,470
The function's only value is $C$, which is the function's absolute maximum value and its absolute minimum value.
 
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Aug 2018
137
7
România
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 a moderator:
Jun 2019
493
262
USA
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 \(\displaystyle \forall x\in \mathbb R\)?!?


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.


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 a moderator:
Aug 2018
137
7
România
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.


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 a moderator:
Oct 2018
129
96
USA
From an Analysis standpoint:

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

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