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

2 Answers
Mar 25, 2015

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

Mar 25, 2015

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