Question

Let f (x) = 1/(x^2) What conclusion can be made about the maximum and minimum values of function f on the interval [–5, 5] using the extreme value theorem?

Answer 1

Answer:

The function is not continuous on the interval, so the extreme value theorem does not apply

Step-by-step explanation:

Recall that the Extreme Value Theorem states that if a function is continuous on a closed interval [a,b], then the function must have a maximum and a minimum on the interval. Given that the function is discontinuous at x=0 which gives off a vertical asymptote, this does not meet the requirements of the theorem.