Hello, Miranda!

#9 is a messy one . . .

9) Find the coordinates of the turning points on the curve with the equation:

. . . y³ + 3xy² - x³ .= .3

Differentiate implicitly: .3y²y'+ 6xyy'+ 3y² - 3x² .= .0

Then we have: .3y²y'+ 6xyy'.= .3y² - 3x²

Factor: .3y(y + 2x)y'.= .3(y² - x²)

. . . . . . . . . . . .y² - x²

Hence: .y'.= .-----------

. . . . . . . . . . .y(y + 2x)

The derivative equals zero when its numerator equals zero:

. . y² - x² .= .0 . → . y = ±x

y = x: .Substitute into the original equation.

. . x³ + 3x·x² - x³ .= .3 . → . x³ = 1 . → . x = 1 . → . y = 1

One turning point is: .(1, 1)

y = -x: .Substitute into the original equation.

. . -x³ + 3x·x² - x³ .= .3 . → . x³ = 3 . → . x = cbrt{3} . → . y = -cbrt{3}

Another turning point is: .(cbrt{3}, -cbrt{3})