Find the minimum value of \(x^2- 3x + 2\) for all real values of x
The correct answer is A. -\(\frac{1}{4}\)
y = x^2- 3x + 2, \(\frac{dy}{dx}\) = 2x - 3
at turning pt, \(\frac{dy}{dx}\) = 0
∴ 2x - 3 = 0
∴ x = \(\frac{3}{2}\)
\(\frac{d^2y}{dx^2}\) = \(\frac{d}{dx}\)(\(\frac{d}{dx}\))
= 270
∴ y
= 2\(\frac{3}{2}\) - 3\(\frac{3}{2}\) + 2
= \(\frac{9}{4}\) - \(\frac{9}{2}\) + 2
= -\(\frac{1}{4}\)
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