Question

How do you find the points where the graph of the function `y'=(-20x^(3)-40x) / (40x + 105y^(6))` has horizontal tangents?

Answer 1

At `x=0`

Explanation:

If `y'=(-20x^(3)-40x) / (40x + 105y^(6))` have horizontal tangents then at those points `y'=0`

so analyzing the solutions of

`(-20x^(3)-40x) = 0 = (20x^2+40)x`

we conclude that the only solution is for `x = 0`