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QUESTION

A function $f(x)$ is defined by

$$f(x)=\frac{x^2-2}{x^2-1}$$

\begin{description}

\item[(i)]
Determine the nature of any stationary points of the function.

\item[(ii)]
Sketch the graph of $f(x)$ showing clearly any stationary points
and asymptotes.

[Note: All the working necessary to obtain your answer must be
clearly shown.]

\end{description}

\bigskip

ANSWER

$\ds f(x)=\frac{x^2-2}{x^2-1}$

\begin{description}

\item[(i)]
$\ds\frac{df}{dx} =\frac{(x^2-1)(2x)-(x^2-2)(2x)}{(x^2-1)^2}
=\frac{2x(x^2-1-x^2+2)}{(x^2-1)^2} =\frac{2x}{(x^2-1)^2}$

Hence there is only one stationary point at $x=0$

\begin{eqnarray*}\frac{d^2f}{dx^2}
&=&\frac{(x^2-1)^22-2x(2(x^2-1)(2x))}{(x^2-1)^4}\\
&=&\frac{(x^2-1)\{2(x^2-1)-8x^2\}}{(x^2-1)^4}\\
&=&\frac{(x^2-1)(-6x^2-2)}{(x^2-1)^4}\\
&=&-2\frac{(1+3x^2)}{(x^2-1)^3}
\end{eqnarray*}

$\ds\left.\frac{d^2f}{dx^2}\right|_{x=0}=-\frac{2}{(-1)^3}=2>0$
therefore $x=0$ is a minimum.

\item[(ii)]
Clearly $f$ becomes undefined when $x^2-1=0,$\ so\ $x=\pm1$ are
the asymptotes. The function is even, since $\ds
f(-x)=\frac{(-x)^2-2}{(-x)^2-1}=\frac{x^2-2}{x^2-1}=f(x)\\
f(0)=\frac{-2}{-1}=2$ therefore the minimum value of the function
$f$ is 2\\ $f(x)=0$ when $x^2=2,\ x=\pm\sqrt{2}$\\ As $\ds
x\to+\infty,\
f(x)=\frac{1-\frac{2}{x^2}}{1-\frac{1}{x^2}}\to\frac{1}{1}=1,$ an
asymptote\\ As $x\to-\infty, f\to 1$ (same as above, or by the
symmetry of even functions)\\
$\ds\frac{d^2f}{dx^2}=-\frac{2(1+3x^2)}{(x^2-1)^3}\neq0$ for any
$x$, therefore there is no point of inflexion.


$\ds x=1^+,\ f=\frac{(-)}{(+)}=(-)\\ x=1^-,\
f=\frac{(-)}{(-)}=(+)\\
f=\frac{x^2-2}{x^2-1}=\frac{x^2-1-1}{x^2-1}=1-\frac{1}{x^2-1}$
therefore $f<1$ as $x\to+\infty$

We can complete the graph since $f$ is even (symmetric about the
$y$-axis).

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