Question: find min and max of f(p)=p^2-7p+6/p-10
Taking the function to be: f(p) = (p^2 - 7p + 6)/(p - 10), or f(p) = (p - 1)(p - 6)/(p - 10)
differentiating f (the first form) gives us,
f' = {(2*p - 7)(p - 10) - (p^2 - 7*p + 6)}/(p - 10)^2 = 0,
(2*p - 7)(p - 10) - (p^2 - 7*p + 6) = 0
2p^2 - 27p + 70 - p^2 + 7p - 6 = 0
p^2 - 20p + 64 = 0
(p - 4)(p - 16) = 0
p = 4, p = 16
f'(p) = (p^2 - 20p + 64)/(p - 10)^2
Differentiating f(p) a second time.
f'' = {(p - 10)^2(2p - 20) - (p^2 - 20p + 64)2(p - 10)} / (p - 10)^4
f'' = 2{(p-10)^2 - p^2 + 20p - 64)} / (p - 10)^3
f'' = 2{(36)} / (p - 10)^3
f'' = 72/(p - 10)^3
There are turning points at p = 4 and at p = 16.
f''(4) = 72/(-6)^2 < 0 => maximum.
f''(16) = 72/(6)^3 > 0 => minimum.
Using the 2nd form of f(p), [f(p) = (p - 1)(p - 6)/(p - 10)]
f(4) = 3*(-2)/(-6) = 1
f(16) = 15*(10)/(6) = 25
Answer: P(4, 1) is a maximum at f(4) = 1. Q(16, 25) is a minimum at f(16) = 25