<s>[INST] Maximizing slope of a secant line Two points on the curve $$ y=\frac{x^3}{1+x^4}$$ have opposite $x$-values, $x$ and $-x$.  Find the points making the slope of the line joining them greatest.
Wouldn't the maximum slope of the secant line be with the max/min of the curve?
So $x=3^{1/4}$ and $x=-3^{1/4}$?
 [/INST] The slope of the line will be
$$slope = \frac{y(x) - y(-x)}{2x} = \frac{2x^3}{2x(1+x^4)} = \frac{x^2}{1+x^4}.$$
Then you take the derivatives of this with respect to $x$ to find the maximum:
$$slope' = \frac{(1+x^4)(2x) - x^2(4x^3)}{(1+x^4)^2} = \frac{2x(1-x^4)}{(1+x^4)^2}.$$
This is zero at $x = 0, -1, and +1$.  You'll find through a second derivative test that the answer is $x = 1$ and $x = -1$.
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