Find the critical points and points of inflection of the curve y=x/(2x-3)^2.

At the point of inflection the curvature of the slope
changes sign.

The function we have is y = x/(2x -
3)^2

y' = x'*(2x - 3)^(-2) + x*[(2x -
3)^(-2)]'

=> (2x - 3)^(-2) + x*(-2)(2x -
3)^(-3)*2

=> (2x - 3)^(-2) - 4x*(2x -
3)^(-3)

=> (2x - 3 - 4x)/(2x -
3)^(3)

=> (-2x - 3)/(2x -
3)^(3)

For x = -3/2, y' =
0

When x < -3/2 , y' is positive and when x >
-3/2 y' is negative.

At the crtical point the value of the
derivative is zero or the function cannot be differentiated. This is true for -2x - 3 =
0 or x = -3/2 and at x = 3/2.

The point of
inflection is at x = -3/2. The critical points are at x = -3/2 and x =
3/2.