Answers

2014-04-23T21:53:03+05:30
Y = 2x^3-9x^2+12x+6
y' = 6x^2-18x+12
for critical point y' = 0
6x^2-18x+12 = 0
(x - 1)(x - 2) = 0
x = 1, 2
y" = 12x-18
for maxima ,minima   y"<0 , y">0
y" = 12*2-18 = 6 > 0    for x=2       (minina)
y" = 12*1-18 = -6 < 0    for x=1       (maxima)
minimum value = 2*2^3-9*2^2+12*2+6
                       = 10
maximum value = 2*1^3-9*1^2+12*1+6
                        =11

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