Find the minimum value of \(x^2- 3x + 2\) for all real values of x

  • A -\(\frac{1}{4}\)
  • B -\(\frac{1}{2}\)
  • C \(\frac{1}{4}\)
  • D \(\frac{1}{2}\)
jamb 1997mathematics

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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