# Constant function

#### Integrator

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

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#### skipjack

Forum Staff
The function's only value is $C$, which is the function's absolute maximum value and its absolute minimum value.

• 1 person

#### Integrator

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

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#### DarnItJimImAnEngineer

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 âˆ€xâˆˆR?
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 $$\displaystyle f(x)=C$$?

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#### Integrator

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 âˆ€xâˆˆR?
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

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#### Greens

From an Analysis standpoint:

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

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