Find the maximum and minimum values, if any, of the following function given by,

$f (x) = 9 x^{2} + 12 x + 2$

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Solution

Given, function is $f (x) = 9 x^{2} + 12 x + 2 = 9 x^{2} + 12 x + 4 - 2$
[we add and subtract 2 for making it perfet square]
$= (9 x^{2} + 6 x + 6 x + 4) - 2 = [3 x (3 x + 2) + 2 (3 x + 2)] - 2 = [(3 x + 2) (3 x + 2)] - 2 = (3 x + 2)^{2} - 2$
It can be observed that $(3 x + 2)^{2} \geq 0 f o r e v e r y x ϵ R$
Therefore, $f (x) = (3 x + 2)^{2} - 2 \geq - 2 f o r e v e r y x ϵ R$
The minimum value of f is attained when $3 x + 2 = 0$
i.e., $3 x + 2 = 0 \Rightarrow x = \frac{- 2}{3}$
$∴$ Minimum value of $f = f (\frac{- 2}{3}) = {(3 \times \frac{- 2}{3} + 2)}^{2} - 2 = - 2$
For any value of x, $f (x) \geq - 2$ , hence function f does not have a particular maximum value.

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