The curve \(y=\left(\dfrac{x-a}{x-b}\right)e^x,\) where \(a\) and \(b\) are constants, has two stationary points. Show that \[a-b<0 \quad \text{ or } \quad a-b>4.\]

Let’s find the stationary points of \[y=\left(\frac{x-a}{x-b}\right)e^x.\]

We plan to differentiate using the product rule here, so it helps us to find the derivative of \(\dfrac{x-a}{x-b}\) first.

If \(z = \dfrac{x-a}{x-b}\), then by the quotient rule, \(z' = \dfrac{a-b}{(x-b)^2}\). Returning to our original problem, we have \(y'=0\) when \[\frac{dy}{dx}=\left(\frac{x-a}{x-b}\right)e^x+\left[\dfrac{a-b}{(x-b)^2}\right]e^x=0\] \[\implies \frac{e^x(x^2-(a+b)x+(a+ab-b))}{(x-b)^2}=0.\]

We know that \(e^x > 0\) for any \(x\), so we need \(x^2-(a+b)x+(a+ab-b)=0.\)

This has two solutions as long as the discriminant is positive, which happens if and only if \((a+b)^2 -4(a+ab-b) > 0.\)

Multiplying out and rearranging here gives \[(a-b)(a-b-4)>0,\] which holds when \[a-b<0 \quad \text{ or } \quad a-b>4.\]

  1. Show that, in the case \(a=0\) and \(b=\dfrac{1}{2}\), there is one stationary point on either side of the curve’s vertical asymptote, and sketch the curve.
The vertical asymptote to the curve is at \(x=\dfrac{1}{2}\). Now let’s substitute in the values of \(a\) and \(b\) to find the stationary points: \[x^2-(a+b)x+ab-b+a=x^2-\frac{1}{2}x-\frac{1}{2}=0.\] Solving this, we find that \[\begin{align*} &x^2-\frac{1}{2}x-\frac{1}{2}=0 \\ &\implies 2x^2-x-1=0\\ &\implies (2x+1)(x-1)=0\\ &\implies x = 1, -\dfrac{1}{2}, &\end{align*}\]

which lie on either side of the asymptote \(x=\dfrac{1}{2}\).

We can now sketch the graph. We know two stationary points, \(\left(-\dfrac{1}{2},\dfrac{e^{-1/2}}{2}\right)\) and \((1,2e)\), and by inspection we can see that it also passes through \((0,0)\).

We also have that \(y\to\infty\) as \(x\to+\infty\), and \(y\to 0\) as \(x\to -\infty\).

Also, as \(x\to \dfrac{1}{2}\) from the right, \(y\to+\infty\) and as \(x \to \dfrac{1}{2}\) from the left \(y \to -\infty\).

We may now give an approximate sketch of the graph.

Graph of the function with the limit, asymptote and turning point behaviour described. The graph has only one root at the origin.

Note that the curve is close to \(y = e^x\) when \(\vert x \vert\) is large.

  1. Sketch the curve in the case \(a=\dfrac{9}{2}\) and \(b=0\).
We again find the stationary points of the curve, this time solving \[\begin{align*} &x^2-\frac{9}{2}x+\frac{9}{2}=0 \\ &\implies 2x^2-9x+9=0\\ &\implies (2x-3)(x-3)=0\\ &\implies x = 3, \dfrac{3}{2}. \end{align*}\]

So our two stationary points are \(\left(\frac{3}{2},-2e^{3/2}\right)\) and \(\left(3,-2e^3\right)\).

The vertical asymptote is at \(x=0\). As \(x\to 0\) from the right, we have that \(y\to-\infty\), since in this case \(x-\tfrac{9}{2} < 0\) and the denominator \(x\) is positive, so \(y\) is negative. As \(x\to 0\) from the left, we have \(y\to +\infty\) as \(x-\tfrac{9}{2} < 0\) again, but this time the denominator \(x\) is negative.

The graph crosses the \(x\)-axis once, at \(x=\dfrac{9}{2}\), and as \(x \to +\infty\), \(y \to +\infty\) and as \(x \to -\infty\), \(y\to 0\).

Graph of the function with behaviour as described.

This applet shows the curve for different values of \(a\) and \(b\).