Can we sketch the curve $y=(x-a)e^x/(x-b)$?

Can we sketch the curve $y=(x-a)e^x/(x-b)$?

Question

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

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.

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1. Sketch the curve in the case $a=\dfrac{9}{2}$ and $b=0$.

Solution

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 $a-b<0 \quad \text{ or } \quad a-b>4.$

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.

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

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 $x \to -\infty$, $y\to 0$.

GeoGebra applet

This applet shows the curve for different values of $a$ and $b$.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/gVxEqSMt/width/627/height/432/border/888888/rc/false/ai/false/sdz/true/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file