Tangent Lines Question 4

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 $\dfrac{1}{2}$ ?

(0,0)

only

\left(\dfrac{1}{2},\dfrac{1}{5}\right)

only

(0,0)

and

(-4,2)

There are no such points.

Approach

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:

x=0

f(0)=\frac{0}{0+2}

(0,0)

x=-4

f(-4)=\frac{-4}{-4+2}

(-4,2)