Question 1052786
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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 = -{{{(64x^2+1)/x}}},  or, which is the same, y = {{{-64x - 1/x}}}.

First derivative is

{{{(dy)/(dx)}}} = {{{-64}}} + {{{1/x^2}}}


The first derivative has a zero at

{{{1/x^2}}} = 64,   or   x = +/-{{{1/8}}}.

There are two stationery points. One is ({{{1/8}}},{{{-16}}}). The other is ({{{-1/8}}},{{{16}}}).

{{{graph( 330, 330, -1.5, 1.5, -100.5, 100.5,
          -64x - 1/x
)}}}


           Plot y = {{{-64x - 1/x}}}

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

On terminology see <A HREF=https://en.wikipedia.org/wiki/Convex_function>this</A> Wikipedia article.
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