Answer :

f(x) = x^{3} – 6x^{2} + 9x + 15

∴ f'(x) = 3x^{2} – 12x + 9 = 3(x^{2} – 4x + 3)

f''(x) = 6x – 12 = 6(x – 2)

for maxima and minima,

f'(x) = 0

3(x^{2} – 4x + 3) = 0

So roots will be x = 3, 1

Now,

f''(3) = 6 > 0

x = 3 is point of local minima

f''(1) = – 6 < 0

x = 1 is point of local maxima

local max value = f(1) = 19

local min value = f(3) = 15

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