Review question

# Can we sketch $y=x^2/(1+x^4)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5501

## Solution

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 $(0,0)$.