Unformatted text preview: g ± ( x ) < 0 when 1 < x < 4 (d) g ±± ( x ) > when 5 /x 2 > 8 /x 3 or x > 8 / 5. 4. Minimize P = x + y + ³ x 2 + y 2 subject to xy = 4 P ( x ) = x + 4 x + ´ x 2 + 16 x 2 P ± ( x ) = 14 x 2 + 1 2 µ x 2 + 16 x 2 ¶1 / 2 µ 2 x32 x 3 ¶ P ± (2) = 0 since when x = 2, (14 /x 2 ) = 0 and 2 x32 x 3 = 0. When x < 2, P ± ( x ) < 0 since (14 /x 2 ) < 0 and 2 x32 x 3 < 0. When x > 2, P ± ( x ) > 0 since (14 /x 2 ) > 0 and 2 x32 x 3 > 0....
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 Spring '07
 DURRETT
 Math, Calculus, Critical Point, dx, 3x2, 5 8 g, R. Durrett

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