Question

Find the absolute maximum & minimum of h(v) = v^2 – 2v + 4 with a closed interval of 0<v<5

Answer 1

h(v)=v²-2v+4, h'(v)=2v-2=0 at extrema, so v=1 and h"(1)=2, so we have a minimum at v=1, which is also the absolute minimum. h(5)=25-10+4=19 and h(0)=4, so the absolute maximum is close to 19, because the upper limit of the interval is <5, according to the given inequality.