Review question

# If $y=(x^2+1)/(x^2-a^2)$, then $y$ cannot take which values? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8765

## Solution

Show that, if $y=\dfrac{x^2+1}{x^2-a^2},$ $y$ takes all real values twice, except those for which $-\dfrac{1}{a^2}\le y\le 1$.

Since $y$ is a function of $x^2$ rather than just $x$ we can show that in general $y(x)=y(-x)$ by substituting and simplifying:

\begin{align*} y(-x)&=\dfrac{(-x)^2+1}{(-x)^2-a^2} \\ &=\dfrac{x^2+1}{x^2-a^2}\\ &=y(x). \end{align*}

This tells us that where $y$ exists and $x \neq -x$ then $y$ values are repeated.

As $x \rightarrow a \quad x^2-a^2 \rightarrow 0$ so $y \rightarrow \pm \infty$.

Since $x^2+1\geq1 \quad y\neq 0$ so there are some values that $y$ cannot take and we must find them.

Rearranging $y=\dfrac{x^2+1}{x^2-a^2}$, we arrive at $x^2=\dfrac{a^2y+1}{y-1}.$

Since $x$ is real $x^2 \geq 0$ so a $y$-value is impossible if and only if $\dfrac{a^2y+1}{y-1} < 0$.

So when is $\dfrac{a^2y+1}{y-1} < 0$?

If $y-1$ is positive, then $a^2y+1<0 \implies y < -\dfrac{1}{a^2}$, which is impossible, since $y > 1$.

On the other hand, if $y-1$ is negative (that is if $y<1$), then $a^2y+1>0 \implies y > -\dfrac{1}{a^2}$, so we have $-\dfrac{1}{a^2}<y<1$.

If $y = 1$, then no $x$-values are possible. If $y = -\dfrac{1}{a^2}$, then $x = 0$, but this $y$-value (occurring at the $y$-intercept) is unique (not repeated) and is the position we noted earlier when $x=-x$.

This means that $y$ takes each real value twice, except when $-\dfrac{1}{a^2}\le y\le 1$.

Sketch the curve $y=\dfrac{x^2+1}{x^2-4}$, indicating its asymptotes.

Since $y(x) = y(-x)$, the graph is symmetrical about the $y$-axis (this is an even function). We have vertical asymptotes at $x = \pm a$.

We can calculate the derivative of $y$: $\frac{dy}{dx}=\frac{2x}{x^2-a^2}-\frac{2x(x^2+1)}{(x^2-a^2)^2}=\frac{2x(x^2-a^2-x^2-1)}{(x^2-a^2)^2}=-\frac{2(a^2+1)x}{(x^2-a^2)^2}.$

We see that $y$ has a stationary point at $x=0$.

Moreover, $y$ is decreasing for positive $x$, and increasing for negative $x$, (except at $\pm a$, where the derivative is undefined).

This tells us that $\left(0, -\dfrac{1}{a^2}\right)$ is a local maximum. Also, we can rewrite $y$ as $y=\frac{x^2+1}{x^2-a^2}=\frac{x^2-a^2+1+a^2}{x^2-a^2}=1+\frac{1+a^2}{x^2-a^2},$ and so $y$ tends to $1$ as $x \to \pm \infty$. This all enables us to sketch the curve $y=\dfrac{x^2+1}{x^2-4}$, as below.