Suppose \(f\) is continous on the finite closed interval \([a,b]\). Then \(f\) has a maximum value and a minimum value on \([a,b]\)

Notes:

• The above theorem allows us an easier way to find the maximum and minimum values of a continous function on a finite closed interval. In the previous sections solution, you could see how difficult is find if something is a local minima or a maxima. But using this method, find the values at the endpoints and the critical numbers - whichever one is the largest must be the maximum value of the function on the interval and whichever one is the smallest is the minimum value on that particular interval.

If \(a \leq u < v \leq b\), then we will say that the interval \([u,v]\) is nonmaximizing if there is some number \(d \in [a,b]\) such that for every \(x \in [u,v]\), \(f(x) < f(d)\).

In other words, the interval \([u,v]\) is nonmaximizing if there is a point \((d,f(d))\) on the graph of \(f\) that is higher than any point on the part of the graph of \(f\) with \(u \leq x \leq v\).

If \([u,v]\) is not nonmaximizing, then we say that it is maximizing.

Notes: If we are looking for a number where \(f\) attains its maximum value on the interval \([a,b]\), there is no point in looking in a nonmaximizing interval.

Suppose that \([u,v]\) is a maximizing interval. Then there are numbers \(u'\) and \(v'\) such that \(u \leq u' < v' \leq v, v' - u' = (v-u)/2\), and \([u',v']\) is maximizing.