Can we sketch $y=x^2/(1+x^4)$?

Can we sketch $y=x^2/(1+x^4)$?

Question

Find the coordinates of the turning points on the curve with equation $y=\frac{x^2}{1+x^4}.$

Sketch the curve.

Suggestion

What do we know about the derivative of a function at a turning point?

Do we know any rules for differentiating functions like this one?

What does the curve look like for large values of $x$?

Solution

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

$(0,0)$

,

$(\pm 1,\dfrac{1}{2})$

.

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 $(0,0)$.