# What is the instantaneous rate of change at x = 2 of the function given by f(x)= (x^2-2)/(x-1)?

Mar 25, 2015

Let $g \left(x\right) = {x}^{2} - 1$ and $h \left(x\right) = x - 1$
so $f \left(x\right) = g \frac{x}{h \left(x\right)}$

By the quotient rule for derivatives
$f ' \left(x\right) = \frac{g ' \left(x\right) \cdot h \left(x\right) - g \left(x\right) \cdot h ' \left(x\right)}{h \left(x\right)} ^ 2$

$g ' \left(x\right) = 2 x$
$h ' \left(x\right) = 1$

So,
$f ' \left(x\right) = \frac{\left(\left(2 x\right) \left(x - 1\right)\right) - \left({x}^{2} - 2\right) \left(1\right)}{x - 1} ^ 2$

$= \frac{{x}^{2} - 2 x + 2}{x - 1} ^ 2$

The instantaneous rate of change at $x = 2$ is therefore

$= \frac{{\left(2\right)}^{2} - 2 \left(2\right) + 1}{\left(2\right) - 1} ^ 2$

$= \frac{2}{1} = 2$

Mar 25, 2015

What is the instantaneous rate of change at $x = 2$ of the function given by $f \left(x\right) = \frac{{x}^{2} - 2}{x - 1}$?

Using the definition:

${\lim}_{x \rightarrow 2} \frac{f \left(x\right) - f \left(2\right)}{x - 2}$

$= {\lim}_{x \rightarrow 2} \frac{\frac{{x}^{2} - 2}{x - 1} - \frac{{2}^{2} - 2}{2 - 1}}{x - 2}$

$= {\lim}_{x \rightarrow 2} \frac{\frac{{x}^{2} - 2}{x - 1} - 2}{x - 2}$

$= {\lim}_{x \rightarrow 2} \frac{\left(\frac{{x}^{2} - 2}{x - 1} - 2\right)}{\left(x - 2\right)} \frac{\left(x - 1\right)}{\left(x - 1\right)}$

=lim_(xrarr2)((x^2-2)-2(x-1))/((x-2)(x-1)

=lim_(xrarr2)(x^2-2-2x+2)/((x-2)(x-1)

=lim_(xrarr2)(x^2-2x)/((x-2)(x-1)

=lim_(xrarr2)(x(x-2))/((x-2)(x-1)

$= {\lim}_{x \rightarrow 2} \frac{x}{x - 1}$

$= \frac{2}{2 - 1} = 2$