Given, f(x) = x3 – 6x2 + 9x + 15
Differentiate with respect to x, we get, f‘(x) = 3x2 – 12x + 9 = 3(x2 – 4x + 3)
= 3 (x – 3) (x – 1)
For all maxima and minima,
f’(x) = 0
= 3(x – 3) (x – 1) = 0
= x = 3, 1
At x = 1, f’(x) changes from positive to negative
Since, x = 1 is a point of Maxima
At x = 3, f‘(x) changes from negative to positive
Since, x = 3 is point of Minima.
Hence, local maxima value f (1) = (1)3 – 6(1)2 + 9(1) + 15 = 19
Local minima value f (3) = (3)3 – 6(3)2 + 9(3) + 15 = 15Answered by Aaryan | 1 year ago
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