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 ( left(\frac{1}{2},\frac{1}{5}
ight) ) only
c ( (0,0) ) and ( (-4,2) )
d ( (0,0) ) and ( left(4,\frac{2}{3}
ight) )

Explanation:

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

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u = x\), \(u^\prime=1\), \(v=x + 2\), \(v^\prime = 1\).

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Step2: Set the derivative equal to \(\frac{1}{2}\)

Since the slope of the tangent line is \(f^\prime(x)\), we set \(\frac{2}{(x + 2)^{2}}=\frac{1}{2}\)
Cross - multiply: \((x + 2)^{2}=4\)
Take the square root of both sides: \(x+2=\pm2\)
Case 1: \(x+2 = 2\), then \(x=0\)
Case 2: \(x + 2=-2\), then \(x=-4\)

Step3: Find the corresponding \(y\) - values

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

Answer:

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