f(x)=2x³-15x²+36x+1
f'(x) = 6x² -30x +36 = 6(x² -5x +6) = 0 => 6(x-3)(x-2) =0 ; x=2,3
x=1,2,3,5
x=1 ; f(1) = 2-15+36+1 = 24
x=2; f(2)= 2x2³ -15x2² +36x2 +1 =29
x=3 ; f(3) = 2x3³ -15x3² + 36x3 +1 = 28
x=5; f(5) = 3x5³ -15x5² + 36x5 +1 =56 -----> കേവല ഉന്നത വില