Probablityyy3

dy/ dxi = 10xi- 2

so xi=1/5

d²y/dxi²=10>0 so we get a minima

So for all xi's equal to 5,

1-2+5 = 4 is the minima

for maxima we have to consider the end points.

So there we should have xi=0 or xi=1

5-2+5 = 8 is the maxima.. (I am not 100% satisfied by the maxima part.. but lets see if this is the answer?)

what is the answer eureka?

may be i am bcos as i said i myself am not 100% convinced...