Answer:
Since the function is not continuous on its entire domain, the Extreme Value Theorem does not apply. Therefore, the absence of an absolute maximum for this function on its domain [0, ∞) does not contradict the Extreme Value Theorem.
Step-by-step explanation:
The Extreme Value Theorem states that if a function is continuous on a closed interval [a, b], it must attain both an absolute maximum and an absolute minimum.
In the case of the function
[tex]\( f(x) = x^2 + \frac{1}{x+1} \)[/tex]
with domain [0, ∞), there's no contradiction with the Extreme Value Theorem because the function is not continuous on its entire domain.
To see why, consider the behavior of the function as x approaches infinity. The term [tex]\( \frac{1}{x+1} \[/tex] approaches zero as x goes to infinity. However, the term [tex]\( x^2 \)[/tex] grows without bounds as x approaches infinity. As a result, the function
[tex]\( f(x) = x^2 + \frac{1}{x+1} \)[/tex]grows without bound as x goes to infinity. This means there is no finite value that the function approaches as x goes to infinity, so the function is not continuous on the interval [0, ∞).
Since the function is not continuous on its entire domain, the Extreme Value Theorem does not apply. Therefore, the absence of an absolute maximum for this function on its domain [0, ∞) does not contradict the Extreme Value Theorem.
