SOLUTION: Find the coordinates of the points on the curve y = x^2/(2x-1) where dy/dx = 0.

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Question 1190935: Find the coordinates of the points on the curve y = x^2/(2x-1) where dy/dx = 0.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
dy/dx=((2x-1)(x^2)'-x^2(2x-1)')/(2x-1)^2

dy%2Fdx=%282x%282x-1%29-2x%5E2%29%2F%282x-1%29%5E2

dy%2Fdx=%282x%5E2-2x%29%2F%282x-1%29%5E2

dy%2Fdx=0=2x%5E2-2x onlyneednumerator

system%28for%2Cx=0%2C+and%2C+cross%28x=1%29%29
point (0,0)

Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.
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. 


%28dy%29%2F%28dx%29 = ((2x-1)(x^2)'-x^2(2x-1)')/(2x-1)^2


%28dy%29%2F%28dx%29 = %282x%2A%282x-1%29-2x%5E2%29%2F%282x-1%29%5E2


%28dy%29%2F%28dx%29 = %282x%5E2-2x%29%2F%282x-1%29%5E2


%28dy%29%2F%28dx%29 = %282x%28x-1%29%29%2F%282x-1%29%5E2


dy%2Fdx = 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).      ANSWER



               Visual illustration


    


    Plot  y = x%5E2%2F%282x-1%29  (red)  and  y= 1  (horizontal tangent, green)

Solved (correctly).