Find the local maxima and local minima, if any of the following function. Also, find the local maximum and the local minimum values, as the case may be as follows.

$f (x) = x^{3} - 6 x^{2} + 9 x + 15$

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Solution

Given functions is, $f (x) = x^{3} - 6 x^{2} + 9 x + 15$
$\Rightarrow f^{'} (x) = 3 x^{2} - 12 x + 9 a n d f " (x) = 6 x - 12$
For maxima or minima put f'(x)=0
$\Rightarrow 3 x^{2} - 12 x + 9 = 0 \Rightarrow 3 (x^{2} - 4 x + 3) = 0 \Rightarrow 3 (x^{2} - 3 x - x + 3) = 0 \Rightarrow 3 [x (x - 3) - 1 (x - 3)] = 0 \Rightarrow 3 (x - 1) (x - 3 = 0 \Rightarrow x = 1 o r x = 3$
At $x = 1 f " (1) = 6 \times 1 - 12 = 6 - 12 = - 6 < 0$
$∴$ x=1 is a point of maxima.
Maximum value $= f (1) = (1)^{3} - 6 (1)^{2} + 9 (1) + 15 = 1 - 6 + 9 + 15 = 19$
At $x = 3, f " (3) = 6 \times 3 - 12 = 18 - 12 = 6 > 0$
x=3 is a point of minima
$∴$ Minimum value $= f (3) = 3^{3} - 6 \times 3^{2} + 9 \times 3 + 15 = 27 - 54 + 27 + 15 = 15$

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