Q. 6
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = x3 – 6x2 + 9x +15
Answer :
We have, f(x) = x3 – 6x2 + 9x + 15
Differentiate w.r.t 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 –ve to + ve
Since, x = – 1 is a point of Maxima
At x =3 f ‘ (x) changes from –ve to + ve
Since, x =3 is point of Minima.
Hence, local max value f(1)= (1)3 – 6(1)2 + 9(1) + 15 = 19
local min value f(3) = (3)3 – 6(3)2 + 9(3) + 15 = 15
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