Review question

# If the gradient is $x(1-x^2)e^{-x^2}$ can we find the stationary points? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6620

## Solution

The curve $C_1$ passes through the origin in the $x\text{-}y$ plane and its gradient is given by $\frac{dy}{dx}=x(1-x^2)e^{-x^2}.$ Show that $C_1$ has a minimum point at the origin and a maximum point at $(1,\frac{1}{2}e^{-1})$. Find the coordinates of the other stationary point. Give a rough sketch of $C_1$.

$\frac{dy}{dx}=x(1-x^2)e^{-x^2}.$ This is zero when $x=0,\pm 1$. To classify these stationary points we need the second derivative: \begin{align*} \frac{d^2y}{dx^2} &= (1-x^2)e^{-x^2}+x(-2x)e^{-x^2}-2x^2(1-x^2)e^{-x^2}\\ &= (1-5x^2+2x^4)e^{-x^2}. \end{align*} Thus \begin{align*} y''(0) &= 1 > 0 \quad \Longrightarrow \quad \text{this is a minimum point,} \\ y''(-1)=y''(1) &= (1-5+2)e^{-1} = -\frac{2}{e} < 0 \quad \Longrightarrow \quad \text{these are maximum points.} \end{align*}

Since we know the curve passes through the origin, $(0,0)$ must be a minimum point.

To find the coordinates of the other stationary points we need to know the values of $y(\pm1)$ which requires us to integrate the given expression. There are (at least) two different ways of doing this.

Our maxima are therefore found at $\left(\pm 1,\dfrac{1}{2e}\right)$. Furthermore, the function is even, that is $f(-x)=f(x)$ for all $x$; and as $x \rightarrow \pm\infty$, $y \rightarrow 0$.

The curve $C_2$ passes through the origin in the $x\text{-}y$ plane and its gradient is given by $\frac{dy}{dx}=x(1-x^2)e^{-x^3}.$ Show that $C_2$ has a minimum point at the origin and a maximum point at $(1,k)$, where $k>\frac{1}{2}e^{-1}$.

So now $\frac{dy}{dx}=x(1-x^2)e^{-x^3}.$

Note the change from $-x^2$ to $-x^3$ in the power of $e$ means this becomes a really hard differential equation to solve.

The methods we used above both fail. Fortunately we can complete the question without directly solving this equation.

The stationary points are again at $x=0,\,\pm 1$. Classifying these: \begin{align*} \frac{d^2y}{dx^2} &= \left(1-x^2\right)e^{-x^3} -2x^2e^{-x^3} -3x^3\left(1-x^2\right)e^{-x^3} \\ &= \left(1-3x^2-3x^3+3x^5\right)e^{-x^3}. \end{align*} Thus \begin{align*} y''(0) &= 1 > 0 \quad \Longrightarrow \quad \text{this is a minimum point,} \\ y''(\pm1) &= (1-3\mp3\pm3)e^{-1} = -\frac{2}{e} < 0 \quad \Longrightarrow \quad \text{these are maximum points.} \end{align*}

Now we need to show the maximum point at $(1,k)$ satisfies $k>\frac{1}{2}e^{-1}$. We cannot evaluate the function directly, but we know that the curve passes through the origin, like $C_1$, and that $C_1$ has a maximum at $(1,\frac{1}{2}e^{-1})$. So we know $C_1$ reaches this point, and we want to show $C_2$ goes above it. For $0<x<1$, we have:

\begin{align*} x^2 &> x^3 \\ \Longrightarrow \quad -x^2 &< -x^3 \\ \Longrightarrow x(1-x^2)e^{-x^2} &< x(1-x^2)e^{-x^3}. \end{align*}

So the derivative of $C_2$ is greater than that of $C_1$ in this range. That means that $C_2$’s $y$ value is “growing” faster than $C_1$’s between the origin and the maximum point at $x=1$, so its maximum point will have greater $y$-coordinate, that is to say, the maximum occurs at $(1,k)$, where

$k>\frac{1}{2}e^{-1}$