Question 1190935
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Find the coordinates of the points on the curve y = x^2/(2x-1) where dy/dx = 0.
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            In this problem, there are TWO points, satisfying the condition  dy/dx=0.


            @josgarithmetic found one point and missed another.


            I came to bring full correct solution.



<pre>
{{{(dy)/(dx)}}} = ((2x-1)(x^2)'-x^2(2x-1)')/(2x-1)^2


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


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


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


{{{dy/dx}}} = 0   has two solutions  x= 0  and  x= 1.


The two points on the curve with zero derivative are  (x,y) = (0,0)  and  (x,y) = (1,1).      <U>ANSWER</U>



               Visual illustration


    {{{graph( 400, 400, -1, 2, -2, 2,      
              x^2/(2x-1), 1
)}}}


    Plot  y = {{{x^2/(2x-1)}}}  (red)  and  y= 1  (horizontal tangent, green)

</pre>

Solved  (correctly).