Question

a) What does the Extreme Value Theorem say?

b) Explain how the Closed Interval Method works.

Short answer

(a) The Extreme value theorem is stated.

(b) A closed interval method is a way to solve a problem within a specific task interval.

Step-by-step solution

Part (a)

The extreme value theorem says that if\(\left( {a,b} \right)\)is a closed and bounded interval and \(f:\left( {a,b} \right) \to R\)be a continuous function defined on \(\left( {a,b} \right)\)then \(f\left( x \right)\)is bounded and it attains its bounded, that is, \(\sup f\left( x \right) \in f\left( {\left( {a,b} \right)} \right)\) here \(x \in \left( {a,b} \right)\) and \(\inf f\left( x \right) \in f\left( {\left( {a,b} \right)} \right)\) here\(x \in \left( {a,b} \right)\).

So \(\exists \alpha ,\beta \in \left( {a,b} \right)\) such that ,

\(f\left( \alpha \right) = \sup f\left( x \right)\)and

\(f\left( \beta \right) = \inf f\left( x \right)\)

According to the extreme value theorem, a function must have a maximum and a minimum on a closed interval (a,b) if it is continuous on the interval.

Part (b)

Suppose f is continuous on some closed, bounded interval\(\left( {a,b} \right)\).

1. Find the critical points\({c_1},{c_2},.....\) of f on \(\left( {a,b} \right)\)
2. Evaluate f at the endpoints a and b the critical points
3. The largest of these values will be the absolute max of f; the smallest will be the absolute min of f on \(\left( {a,b} \right)\).

Where, the graph for the function is of the form.