Review question

# Can we show the function $(2x+1)/(x^2-1)$ can take all real values? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9329

## Solution

Prove that, for real values of $x$, the function $\frac{2x+1}{x^2-1}$ can take all real values.

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.

Sketch the graph of the function, and prove that it has a point of inflection between $x=-1$ and $x=+1$.

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.

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 $0.5$.