Maximum and Minimum Limits at Infinity Proof?

[Nebula]

I'm frustrated beyond belief with a proof.

Suppose we have an continuous even function with a domain of all real numbers. Now this function has limit as x goes to negative infinty equal to l and the limit as x goes to positive infinty is also equal to l.

I want to show that this function will either have a maximum or a minimum.

I'm not sure at all how to show this rigorously since I don't know how to apply the definition of a limit to limits at infinity. I think it has to do with bounds. And I need to do this without first derivative test.

 

[selfAdjoint]

Aren't the specifications you give consistent with the constant function f(x) = 1?

 

[Nebula]

Did you think my l was a 1. Maybe I should write it differently.

lim (x-> -oo) f(x) = lim (x-> oo) f(x) = a
and a is even.

Want to show f has either a minimum or a maximum.

 

[T@P]

intuitively this makes sense, but rigorously you could show that unless it is a consant function (for example y = 1) then there must be a point where it switches between a positive and negative slope. I am not entirely sure what level of "rigorousness" you want.

 

[HallsofIvy]

First, you still have the problem that was pointed out by both selfadjoint and T@p:
The constant function f(x)= a satisfies your conditions but does not have a maximum or minimum so the "theorem" as you stated it is not true.

If f(x) is NOT A CONSTANT FUNCTION, then there exist some x₀ such that f(x₀) is not equal to a and so is either larger than or less than a.

Assume f(x₀)> a. Since limit as x-> infinity f(x)= a, there exist some x₁> x₀ such that f(x₁)< f(x₀).
Similarly, since limit as x-> -infinity f(x)= a, there exist some x₂< x₀ such that f(x₂)< f(x₀). Since f is continuous on the closed and bounded interval [x₂, x₁] it must have both maximum and minimum values there. Now show that f has a maximum on -infinity to infinity.

 

[jacksondr]

I need help. Find the minimum of y = Absolute value of (sinx + cosx + tanx + cotx +
secx + cscx) Thanks Ruth Jackson the_perfect_mom@hotmail.com

 

[shmoe]

Assume f(x₀)> a. Since limit as x-> infinity f(x)= a, there exist some x₁> x₀ such that f(x₁)< f(x₀).

I think you need to add something about x₁ here. Choose it in such a way that if x>x₁ then f(x)<f(x₀). Similar change with how you select x₂.