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 $x=-3^{1/4}$?
 A: 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 $x = -1$.
A: Replace $x^2=\tan(\alpha)$. We have $$\text{Slope} = \frac{\frac{x^3}{1+x^4}-\frac{-x^3}{1+x^4}}{2x}= \frac{x^2}{1+x^4} = \frac{\sin(2\alpha)}{2}\le\frac{1}{2}$$ with equality exactly when $\alpha=n\pi+\frac{\pi}{4}, x^2=1$
A: Hints
Let 
$$y(x)=\frac{x^3}{1+x^4}; $$
so $y(-x)=-y(x)$. We take any $2$ points $P=(x,y(x))$ and $Q=(-x,-y(x))$ like in the OP.
The line through them satisfies
$$\frac{y(x)+y(x)}{x+x}=m; $$
in fact, the line between 2 points $(x_1,y_1)$ and $(x_2,y_2)$ in the $\mathbb R^2$ plane is determined by the equation
$$\frac{y_2-y_1}{x_2-x_1}=s, $$
where $s$ is the slope of the line itself.
To solve the problem we need to maximize $m(x)=\frac{x^2}{1+x^4}$, with $x\in\mathbb R$.
Can you arrive at the extrema $x=0,\pm 1$ and finish the proof?
A: 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.
