difficult differentiation (1 Viewer)

Kaatie · Aug 11, 2009

if y=x^(2n), where n is a positive interger
i) find any stationary pts and determine their nature
ii) show the function has no inflexion pts

cheers

gurmies · Aug 11, 2009

$y = x^{2n} \\\\ \frac{dy}{dx}=2n(x^{2n-1}) \\\\ \frac{d^{2}y}{dx^{2}}=2n(2n-1)(x^{2n-2}) \\\\ \ At \,\, stationary \,\, points, \,\, \frac{dy}{dx}=0 \\\\ 2n(x^{2n-1})=0 \\\\ x = 0, \,\, y=0,\,\, \frac{d^{2}y}{dx^{2}}=0 \\\\ When \,\, x = 1, \,\, \frac{dy}{dx}>0 \\\\ When x = -1, \,\, \frac{dy}{dx}<0 \\\\ Note \,\, that \,\, y = x^{2n} \,\, is \,\, equivalent \,\, to \,\, y=(x^{2})^{n}. \\\\ Since \,\, n \,\, is \,\, a \,\, positive \,\, integer, \,\, it \,\, will \,\, always \,\, {\color{red} resemble} \,\, a \,\, parabola \,\, which \,\, has \,\, a \,\, minimum \,\, at \,\, (0,0) \,\, and \,\, no \,\, points \,\, of \,\, inflexion.$

God-dammit for some reason half my post was left out. Anyway, you conclude that (0,0) is a minimum point through testing on either side. When you let the double derivative = 0, you find that it yields x = 0. This cannot be an inflexion point because concavity doesn't change, PLUS we've determined that (0,0) is a mimimum point.

difficult differentiation (1 Viewer)

Kaatie

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gurmies

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