Find the coordinates of the turning points on the curve with equation \[y=\frac{x^2}{1+x^4}.\]
Finding the derivative \(y'\) using the product rule with \(y=x^2\dfrac{1}{1+x^4}\): \[\frac{dy}{dx}=\frac{2x}{1+x^4}-\frac{4x^3\times x^2}{(1+x^4)^2}=\frac{2x(1+x^4)-4x^5}{(1+x^4)^2}.\] So when \(y'=0\),
\[\begin{align*}
2x(1+x^4) - 4x^5 &= 0 \\
\Longrightarrow 2x-2x^5 &= 0 \\
\Longrightarrow x(1-x^4) &= 0 \\
\Longrightarrow x(1-x^2)(x^2+1) &=0\\
\quad \Longrightarrow \quad x=0, \,\, x=\pm 1.
\end{align*}\]
Substituting back into the equation we find that the coordinates of the turning points are \((0,0)\), \((\pm 1,\dfrac{1}{2})\).
Sketch the curve.
For large \(x\), the \(x^4\) term dominates, and so \(y \rightarrow 0 \quad \text{as} \quad |x| \rightarrow \infty\).
The only root coincides with the stationary point \((0,0)\).
