Question
Determine the point on the curve y=x^2(x^3-x)^2 where the tangent line is horizontal.
Answers
Reiny
I would change the function to
y = x^4(x^2-1)^2, then use the product rule to get
dy/dx = 4(2)(x2-1)(2x) + 4x^3(x^2-1)^2
= 4x^3(x^2-1)(2x^2 - 1)
setting this equal to zero
x = 0 or x = ±1 or x = ±1/√2
plug those x values back into the original to the get their y values
y = x^4(x^2-1)^2, then use the product rule to get
dy/dx = 4(2)(x2-1)(2x) + 4x^3(x^2-1)^2
= 4x^3(x^2-1)(2x^2 - 1)
setting this equal to zero
x = 0 or x = ±1 or x = ±1/√2
plug those x values back into the original to the get their y values