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 · 2016-01-06T15:16:55

Differentiate using the quotient rule and then use the value of x to get $-2$

The rate of change is the derivative of the function. To find this we use the quotient rule $d/dx(f(x)/g(x)) =(g(x)f'(x) - g'(x)f(x))/(g(x)^2)$

$d/dx(f(x)/g(x)) = ((x-1)*2x - 1*(x^2 -2))/(x-1)^2$

$d/dx(f(x)/g(x)) = (2x^2 - 2x -x^2 +2)/(x-1)^2 = (x^2 - 2x - 2)/(x-1)^2$

If $x=2$ the rate of change $(2^2 -2*2-2)/(2-1)^2 =-2/1 = -2$

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?