How do you find the instantaneous rate of change of the function $f(x)=x/(x+2)$ when x=2?

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Answer 1 · 2016-02-07T21:32:04

$f'(2)=1/8$ , so the instantaneous rate of change at $x=2$ is $1/8$

The instantaneous rate of change at $x=2$ is equal to the value of the derivative at $x=2$ .

To find the derivative, use the quotient rule, which states that

$d/dx[g(x)/(h(x))]=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2$

Applying this to the function at hand, we see that

$f'(x)=((x+2)d/dx[x]-xd/dx[x+2])/(x+2)^2$

Note that both of these derivatives are equal to $1$ .

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

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

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

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

$f'(2)=2/(2+2)^2=2/4^2=2/16=1/8$

How do you find the instantaneous rate of change of the function f(x)=x/(x+2)f(x)=x/(x+2) when x=2?