SOLUTION: Find the nature of the stationery points of the curve xy+ 64x^2+1=0 and the coordinates of these points

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Question 1052786: Find the nature of the stationery points of the curve xy+ 64x^2+1=0 and the coordinates of these points
Answer by ikleyn(52790) About Me  (Show Source):
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Find the nature of the stationery points of the curve xy+ 64x^2+1=0 and the coordinates of these points
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Your curve is

y = -%2864x%5E2%2B1%29%2Fx,  or, which is the same, y = -64x+-+1%2Fx.

First derivative is

%28dy%29%2F%28dx%29 = -64 + 1%2Fx%5E2


The first derivative has a zero at

1%2Fx%5E2 = 64,   or   x = +/-1%2F8.

There are two stationery points. One is (1%2F8,-16). The other is (-1%2F8,16).




           Plot y = -64x+-+1%2Fx

The function is an odd function and has two branches.
The branch in the second quarter is convex downward.
The branch in the fourth quarter is convex upward.

On terminology see this Wikipedia article.