Question

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

Answer 1

Let `g(x) = x^2 - 1` and `h(x) = x-1`
so `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

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`