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 June 28th, 2019, 09:08 PM #1 Member   Joined: Aug 2018 From: România Posts: 88 Thanks: 6 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. June 28th, 2019, 10:58 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2218 The function's only value is $C$, which is the function's absolute maximum value and its absolute minimum value. Thanks from Integrator June 29th, 2019, 09:29 PM   #3
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Quote:
 Originally Posted by skipjack 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. June 30th, 2019, 08:25 AM   #4
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Quote:
 Originally Posted by Integrator 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.

Quote:
 Originally Posted by Integrator 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. June 30th, 2019, 08:45 PM   #5
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Quote:
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 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. June 30th, 2019, 09:09 PM #6 Member   Joined: Oct 2018 From: USA Posts: 90 Thanks: 61 Math Focus: Algebraic Geometry From an Analysis standpoint: $\displaystyle f: \mathbb{R} \mapsto \{C\}$ max$\{C\}$ = min$\{C\} = C$ Tags constant, function Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ag07 Complex Analysis 6 December 4th, 2015 07:05 PM mathbalarka Number Theory 21 June 18th, 2012 05:41 PM snjvsingh Complex Analysis 1 April 20th, 2011 09:33 AM Simke Calculus 4 April 6th, 2011 12:06 PM fortin946 Number Theory 4 March 16th, 2011 06:05 PM

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