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 $x=-3^{1/4}$?

Let $m$ be the slope. Then, 
$$m = \frac{y(x) - y(-x)}{2x} = \frac{2x^3}{2x(1+x^4)} = \frac{x^2}{1+x^4}.$$
Find $m'(x)$ and set this to $0$. 
Well, to find $m'(x)$, we use the Quotient Rule and we have $$ m'(x) = \dfrac {(1+x^4)(2x) - (x^2)(4x^3)}{(1+x^4)^2} = \dfrac {2x^5+2x-4x^5=2x-2x^5}{(1+x^4)^2}. $$Well, so the numerator must be $0$, so we have $$ 2x - 2x^5 = 0 \implies x \cdot (1-x^4) = 0. $$Thus, the only real solutions for $x$ are $x=0$ and $x=\pm1$. Finish up from here.
