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 \((0,0)\).

Graph required in the question. The graph is symmetrical in the y axis, it has a minimum at the origin and maxima at x = plus or minus 1. It then decays toward 0 as x tends to plus or minus infinity.