Question

the function f is defined by $f(x)=\frac{x}{x + 2}$. what points $(x,y)$ on the graph of f have the property that the line tangent to f at $(x,y)$ has slope $\frac{1}{2}$?
(a) $(0,0)$ only
(b) $(\frac{1}{2},\frac{1}{5})$ only
(c) $(0,0)$ and $(-4,2)$
(d) $(0,0)$ and $(4,\frac{2}{3})$
(e) there are no such points.

Explanation:

Step1: Find the derivative of $f(x)$

Use the quotient - rule. If $f(x)=\frac{u(x)}{v(x)}$ where $u(x) = x$ and $v(x)=x + 2$, then $f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^2}$. Here, $u^\prime(x)=1$ and $v^\prime(x)=1$. So $f^\prime(x)=\frac{1\cdot(x + 2)-x\cdot1}{(x + 2)^2}=\frac{x+2 - x}{(x + 2)^2}=\frac{2}{(x + 2)^2}$.

Step2: Set the derivative equal to the given slope

We want $f^\prime(x)=\frac{1}{2}$. So $\frac{2}{(x + 2)^2}=\frac{1}{2}$. Cross - multiply to get $(x + 2)^2=4$.

Step3: Solve for $x$

Take the square root of both sides: $x + 2=\pm2$.

• Case 1: If $x+2 = 2$, then $x=0$.
• Case 2: If $x + 2=-2$, then $x=-4$.

Step4: Find the corresponding $y$ - values

• When $x = 0$, $y=f(0)=\frac{0}{0 + 2}=0$.
• When $x=-4$, $y=f(-4)=\frac{-4}{-4 + 2}=2$.

Answer:

C. $(0,0)$ and $(-4,2)$