The function \(f\) is defined by \(f(x)=\dfrac{x}{x+2}\). What points \((x,y) \) on the graph of \(f\) have the property that the line tangent to \(f\) at \((x,y)\) has slope \(\dfrac{1}{2}\)?

Summary Submit Skip Question

$$ f(x)=\frac{x}{x+2} $$ $$ f'(x)=\frac{2}{(x+2)^2} $$

Setting the derivative equal to \(\dfrac{1}{2}\):

$$ \frac{1}{2}=\frac{2}{(x+2)^2} $$ $$ (x+2)^2=4 $$ $$ x=0, -4 $$

Obtaining the \(y\) values: