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June 28th, 2019, 09:08 PM  #1 
Member Joined: Aug 2018 From: RomÃ¢nia 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.

June 29th, 2019, 09:29 PM  #3  
Member Joined: Aug 2018 From: RomÃ¢nia Posts: 88 Thanks: 6  Quote:
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  
Senior Member Joined: Jun 2019 From: USA Posts: 196 Thanks: 78  Quote:
$\displaystyle f'(x)= 0$ 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:
Would it help if we called $C$ the nonlocalised global minimum and the nonlocalised 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  
Member Joined: Aug 2018 From: RomÃ¢nia Posts: 88 Thanks: 6  Quote:
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$\{C\} = C$ 

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