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

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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.

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