Why is continuity necessary before applying the Extreme Value Theorem?

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

Why cannot we say that $f(2.999999999) ≥ f(x)$ and therefore absolute max at f(2.99999999999999) (without reasoning from the extreme value theorem)?

Many thanks!

 

[fresh_42]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 326186
Why cannot we say that $f(2.999999999) ≥ f(x)$ and therefore absolute max at f(2.99999999999999) (without reasoning from the extreme value theorem)?

Many thanks!

We get always a higher value with every 9 we add.

3 is a supremum (lowest higher boundary) of the function, but not a maximum (highest value).

 

[Mark44]

Why cannot we say that $f(2.999999999) ≥ f(x)$ and therefore absolute max at f(2.99999999999999) (without reasoning from the extreme value theorem)?

You are confusing the function value with a value in the domain. 2.9999999 is not even in the domain.

With that correction, your question becomes "Why cannot we say that $f(0.999999999) ≥ f(x)$?
Note that there are values of x closer to 1 than 0.99999999, so for those x-values, f(x) > f(0999999999).
In case you aren't aware, the extreme value theorem doesn't apply here, as far as a maximum value is concerned. The necessary condition is that the function must be continuous on a closed bounded interval. The graph of the left branch of the function is continuous only on the half-open interval [0, 1) and is discontinuous at x = 1.

 

[FactChecker]

What part of the problem states that the Extreme Value Theorem is required to prove the problem statements? This example shows that continuity is required before the Extreme Value Theorem can be applied. I suspect that you have misunderstood the point of the example.