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Show that xx is minimum at x = 1/e |
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Answer» y = xx dy/dx = xx(1 +log x) For minima dy/dx = 0 xx(1 + logx) = 0 ⇒ x = \(\frac{1}{e}\) dy/dx|e = 1/e = (1/e)1/e{e + 0) > 0 ∴ y is minimum at x = \(\frac{1}{e}\) |
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