,
ed hardy canada
Fine a class analysis of a familiar problem
osa'···,
GHD Glätteisen günstig,( a) of fl (a) + and ,
Salvatore Ferragamo outlet, 2 (a) and . * . One 0 directly from the basic inequality was , (a) +2 ≥ 2 √ 2:2, namely , (a) a = 2 . This is clearly wrong. In fact, =. Have 0062a: 4, so . Does not exist, and thus the equal sign to get any Z0) sa Solution 2 : Using the monotonicity of the function solve . Order 0060 '= t, then 0 0 . ' . g (t1)> g (t2), so the plant (f) in (0,1 ] is monotone decreasing , so g (f) = g (1) = 1 +2 = guide , that f ( mouth ) a = number. solution 3: use of discriminant method for solving . COS2 original style into the mouth a 2ycosa +4 = 0, so oOs port = t, then 0 0 was ≥ 2, to determine the plant (t) rain = = 2 the answer is obviously wrong , because the equation △ ≥ 0 can only guarantee t- 2yt +4 = 0 has real roots , but can not guarantee that in fact the root in ( 0,1] and, therefore, can not guarantee that equation o . sza a 2ycosa +4 = 0 to solve. solution 4 : Use the number of shape migration solution by the meaning of the questions , plant (a) + a oc ~ 2a +4 bis ( 12cma2cma 10 ' is easy to know can be seen as fixed points P (2cc ~,) to the point O (o, a 4 ) connection of the ramp Zo0S mouth of a U rate by ( at <-g) can get Y = 1X2 (0 <≤ 2). that the point P trajectory is a parabola ( Figure ) from the image is easy to see , when the point P at the endpoint ( 2,1) when , PQ slope of the minimum , it works (a) = second base 2:2-02 'P a 0/xn
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