How do you find the instantaneous rate of change for $f(x)=(x^2-2)/(x-1)$ for x=2?

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Answer 1 · 2018-02-02T14:59:00

Please refer to the explanation below.

We have to take the derivative of the function $f(x)$ , and then plug in $x=2$ into $f'(x)$ .

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

So, we have to use the quotient rule, which states that,

$d/dx(u/v)=(u'v-uv')/v^2$

Here, $u=x^2-2$ , $v=x-1$

$:.f'(x)=(2x(x-1)-1*(x^2-2))/(x-1)^2$

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

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

Plugging in $x=2$ , we get

$f'(2)=(2^2-2*2+2)/(2-1)^2$

$f'(2)=(4-4+2)/1^2$

$f'(2)=2/1$

$f'(2)=2$

So, the instantaneous rate of change will be $2$ .

Answer 2 · 2018-02-02T15:01:42

$m=2$

By finding the derivativ of $f(x)$

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

Now we determinate the value of the divertive at the point x=2

$f'(2)=2$

How do you find the instantaneous rate of change for f(x)=(x^2-2)/(x-1)f(x)=(x^2-2)/(x-1) for x=2?