Answer :

f (x) = x3 – 6x2 + 9x + 15

f’(x) = 3x2 – 12x + 9


Now, f’(x) = 0


3x2 – 12x + 9 = 0


3(x-1)(x-3) = 0


x = 1,3


g’’(x) = 6x – 12 =6(x-2)


Now, f’(1) = 6(1-2)=-6 < 0


and f’(3) = 6(3-2) = 6 > 0


Then, by second derivative test,


x = 1 is point of local maxima and local maximum of f at x = 1 is


f(1) = 13 – 6 +9 +15 = 19


And,


x = 3 is point of local minima and local minimum value of f at x = 3 is


f(3) = 27 – 54 + 27 +15 = 15


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