Given f (x) = x3 – 6x2 + 9x + 15
Differentiating f with respect to x
∴ f'(x) = 3x2 – 12x + 9 = 3(x2 – 4x + 3)
f” (x) = 6x – 12 = 6(x – 2)
For maxima and minima, f'(x) = 0
3(x2 – 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 and local min value = f (3) = 15
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