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

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Answer 1 · 2015-03-25T18:41:28

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

By the quotient rule for derivatives
$f'(x) = (g'(x)*h(x) - g(x)*h'(x))/(h(x))^2$

$g'(x) = 2x$
$h'(x) = 1$

So,
$f'(x) = (((2x)(x-1)) - (x^2-2)(1))/(x-1)^2$

$= (x^2 - 2x + 2)/(x-1)^2$

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

$= ((2)^2 - 2(2) + 1)/((2)-1)^2$

$= 2/1=2$

Answer 2 · 2015-03-25T21:14:25

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

Using the definition:

$lim_(xrarr2)(f(x)-f(2))/(x-2)$

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

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

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

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

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

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

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

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

$=2/(2-1)=2$

What is the instantaneous rate of change at x = 2x=2x = 2 of the function given by f(x)= (x^2-2)/(x-1)f(x)=x2−2x−1f(x)= (x^2-2)/(x-1)?