If \(f(x) = 2x^{2} - 3x - 1\), find the value of x for which f(x) is minimum.
\(y = 2x^{2} - 3x - 1\)
\(\frac{\mathrm d y}{\mathrm d x} = 4x - 3 = 0\) (At turning point)
\(4x = 3 \implies x = \frac{3}{4}\)
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