1. Show that the curve \(y=x(5-x)^4\) has a turning point where \(x=5\). Determine whether this turning point is a maximum or a minimum.

Using the Product Rule, we have that

\[\begin{align*} \frac{dy}{dx}&=(5-x)^4-4x(5-x)^3\\ &=(5-x)^3[5-x-4x] \\ &=5(5-x)^3(1-x). \end{align*}\]

We see that \(\dfrac{dy}{dx}\) vanishes when \(x=5\), therefore \((5,0)\) is a turning point.

We can see from the expression for \(\dfrac{dy}{dx}\) that \(\dfrac{d^2y}{dx^2}\) will also vanish at \((5,0)\), so calculating this won’t help decide if we have a maximum or a minimum.

Let’s instead look at the expression for \(y\).

We know that when \(x=5\), \(y=0\).

When \(x\) is slightly smaller than \(5\), \(y\) is positive, and the same is true when \(x\) is slightly larger than \(5\), as \(x\) and \((5-x)^4\) are both positive.

Therefore \((5,0)\) is a local minimum for \(y\).

  1. It is given that that \(\dfrac{dy}{dx}\) is inversely proportional to \(x^2\) and that \(y\) and \(\dfrac{dy}{dx}\) are each equal to \(1\) when \(x=2\). Express \(y\) in terms of \(x\).

If \(\dfrac{dy}{dx}\) is inversely proportional to \(x^2\), then we can write \[\frac{dy}{dx}=\frac{c}{x^2}.\] Since \(\dfrac{dy}{dx}=1\) when \(x=2\), we see that \(c=4\), and so \[\frac{dy}{dx}=\frac{4}{x^2}.\] We now integrate this, and see that \[y=-\frac{4}{x}+d.\] Using that \(y=1\) when \(x=2\), we see that \(d=3\).

Therefore \(y=3-\dfrac{4}{x}\).