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

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Answer 1 · 2016-07-21T10:34:30

At $x=0$

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 $x = 0$

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