Find the minimum or maximum value of $f (x) = 2 x^{2} + 3 x - 5$

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Solution

The function $f (x) = 2 x^{2} + 3 x - 5$ is a quadratic function with $a = 2$ and $b = 3$ .

Thus, the maximum or minimum value occurs at:

$x = - \frac{b}{2 a} = - \frac{3}{2 (2)} = - \frac{3}{4}$

Since $a > 0$ , the function has the minimum value and that is:

$f (- \frac{3}{4}) = 2 {(- \frac{3}{4})}^{2} + 3 (- \frac{3}{4}) - 5 = 2 (\frac{9}{16}) - \frac{9}{4} - 5 = \frac{9}{8} - \frac{9}{4} - 5 = \frac{9 - 18 - 40}{8} = - \frac{49}{8}$ .

Hence, the minimum value of the given function is $- \frac{49}{8}$ .

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