Floppy hair

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To find the coordinates of the maximum of each curve, we can differentiate each equation. This gives us a formula for the gradient and this is zero at the maximum.

To make this easier to differentiate, we multiply out the bracket first. \[\begin{align*} y &= 3x-x^2\\ \frac{dy}{dx} &= 3-2x \end{align*}\]

Setting \(\frac{dy}{dx}=0\) gives us \(3-2x=0\) and hence \(x=\frac{3}{2}\). The black curve, \(e\), appears to have a maximum at \(x=1.5\), suggesting that this is the equation of \(e\).

We start by writing the square root in index form and multiplying out the bracket. \[\begin{align*} y &= x^{\frac{1}{2}}(3-x)\\ &= 3x^{\frac{1}{2}} - x^{\frac{3}{2}}\\ \frac{dy}{dx} &= \frac{3}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} \end{align*}\] So the maximum is when \[\begin{align*} \frac{3}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} &= 0\\ x^{-\frac{1}{2}} &= x^{\frac{1}{2}}\\ x &= 1 \end{align*}\]

So this is the green curve, \(c\).

As before, we start by multiplying out the brackets, then differentiate to find the gradient. \[\begin{align*} y &= \frac{1}{4}\left( x^3-8x^2+15x \right)\\ \frac{dy}{dx} &= \frac{1}{4}\left( 3x^2-16x+15 \right) \end{align*}\]

Setting the gradient equal to zero gives us a quadratic equation with roots \(x=\dfrac{8}{3}\pm\dfrac{\sqrt{19}}{3}\). The one of these roots that is between zero and three is at \(x\approx1.21\), so this appears to be the purple curve, \(d\).

\[\begin{align*} y &= 3x^{\frac{1}{3}} - x^{\frac{4}{3}}\\ \frac{dy}{dx} &= x^{-\frac{2}{3}} - \frac{4}{3}x^{\frac{1}{3}} \end{align*}\] At the maximum, \[\begin{align*} x^{-\frac{2}{3}} - \frac{4}{3}x^{\frac{1}{3}} &= 0\\ x^{-\frac{2}{3}} &= \frac{4}{3}x^{\frac{1}{3}}\\ \frac{3}{4} &= x \end{align*}\]

So this appears to be the blue curve, \(b\).

\[\begin{align*} y &= \frac{1}{6}\left( -x^3-2x^2+15x \right)\\ \frac{dy}{dx} &= \frac{1}{6}\left( -3x^2-4x+15 \right) \end{align*}\]

As above, we have a quadratic equation which factorises to give \(x=-3\) or \(\frac{5}{3}\). The black curve, \(e\), has a maximum close to \(x=\frac{5}{3}\), but we have already suggested the first equation for this curve. The first equation does look a better fit, but to check we could calculate the \(y\) values of the two candidates at their maxima and see which ones matches.

The first equation at \(x=\frac{3}{2}\) gives \(y=\frac{9}{4}=2.25\).

This equation at \(x=\frac{5}{3}\) gives \(y=\frac{200}{81}\approx2.5\).

The first equation is definitely a better match for curve \(e\), so this equation is the one without a graph.

By elimination, this must be the remaining curve, \(a\), but let’s do the calculation to be sure.

\[\begin{align*} y &= \frac{2}{\sqrt[3] 9 -1} \left( \sqrt[3]{9}x^\frac{1}{3} -x \right)\\ \frac{dy}{dx} &= \frac{2}{\sqrt[3] 9 -1} \left( \frac{\sqrt[3]{9}}{3}x^{-\frac{2}{3}} -1 \right) \end{align*}\] So at the maximum, \[\begin{align*} \frac{\sqrt[3]{9}}{3}x^{-\frac{2}{3}} -1 &= 0\\ x^{\frac{2}{3}} &= \frac{\sqrt[3]{9}}{3}\\ x^2 &= \frac{9}{3^3} = \frac{1}{3}\\ x &= \pm \frac{1}{\sqrt{3}} \approx \pm 0.58 \end{align*}\]

This does indeed match graph \(a\).