Can we show the function $(2x+1)/(x^2-1)$ can take all real values?

Can we show the function $(2x+1)/(x^2-1)$ can take all real values?

Question

Prove that, for real values of $x$, the function $\frac{2x+1}{x^2-1}$ can take all real values. Sketch the graph of the function, and prove that it has a point of inflection between $x=-1$ and $x=+1$.

Suggestion

We want to show that as we travel from $-\infty$ to $\infty$ along the $x$-axis, $f(x)$ at some point takes each real value.

Suppose that $y=\dfrac{2x+1}{x^2-1}$. Given a $y$, can we find an $x$?

Things to look out for when curve sketching:

• asymptotes;

• intercepts;

• stationary points and their nature;

• behaviour as $x\rightarrow \pm \infty$.

What is the definition of a point of inflection?

Solution

Let

$y=\dfrac{2x+1}{x^{2}-1}$

(

$x \neq \pm 1$

). Rearranging, we have that

$$yx^2-2x-(1+y)=0.$$

We can solve this quadratic for

$x$

to find that (if

$y \neq 0$

),

$$\begin{align*} x &= \frac{2\pm\sqrt{4+4y(y+1)}}{2y}\\ &= \frac{1 \pm \sqrt{y^2 + y + 1}}{y}\\ &=\frac{1 \pm \sqrt{\left(y+\frac{1}{2}\right)^2 + \frac{3}{4}}}{y}. \end{align*}$$

So for any value of $y\neq 0$, this gives real values for $x$, and so for these $x$-values, $f(x)=\frac{2x+1}{x^2-1} = y.$

If

$y = 0$

, then

$x = -\dfrac{1}{2}$

gives

$f(x) = y$

. So the function takes all real values.

There are vertical asymptotes at $x=\pm 1$, and the curve behaves like $y=\dfrac{2}{x}$ for large $x$.

Also $f(0)=-1$ and the only root for $f(x) = 0$ is $-\frac{1}{2}$.

We can find the gradient of the curve by using the

quotient rule

(

$x \neq \pm 1$

):

$$\begin{align*} f'(x) &= \frac{2(x^2-1)-(2x+1)(2x)}{(x^2-1)^2}\\ &= \frac{-2x^2-2x-2}{(x^2-1)^2}\\ &= \frac{-2(x^2+x+1)}{(x^2-1)^2}\\ &= \frac{-2\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)}{(x^2-1)^2},\\ \end{align*}$$

which is always negative.

A point of inflection on a curve occurs where the gradient function has a maximum or a minimum.

As a point moves around the curve, a point of inflection happens where the tangent at that point changes from being on one side of the curve to the other.

Informally, our curve must change from being concave upwards to concave downwards in between the two asymptotes $x\pm1$, and so it has a point of inflection somewhere there.

To be sure of this, we note that $f'(-0.5) = -\dfrac{24}{9}$, $f'(0) = -2$, $f'(0.5) = -\dfrac{56}{9}$, and that $-\dfrac{24}{9} < -2 > -\dfrac{56}{9}$.

This means, if we imagine the graph of $y=f'(x)$, a continuous function between $-1$ and $1$, that $f'(x)$ must have a local maximum between $-0.5$ and $0.5$.