We need to find the derivative first. Think of x/(x+2) as x(x+2)^(-1). That will allow us to use the product rule. You could also use the quotient rule if you want. I'll go with the product rule route.
[tex]f(x) = \frac{x}{x+2}\\\\f(x) = x(x+2)^{-1}\\\\f'(x) = (x+2)^{-1}-x(x+2)^{-2}\\\\f'(x) = \frac{1}{x+2}-\frac{x}{(x+2)^2}\\\\f'(x) = \frac{x+2}{(x+2)^2}-\frac{x}{(x+2)^2}\\\\f'(x) = \frac{x+2-x}{(x+2)^2}\\\\f'(x) = \frac{2}{(x+2)^2}\\\\f'(x) = \frac{2}{x^2+4x+4}\\\\[/tex]
Since we want the tangent slope to be 1/2, we'll set f ' (x) equal to 1/2 and solve for x.
[tex]f'(x) = \frac{2}{x^2+4x+4}\\\\\frac{1}{2} = \frac{2}{x^2+4x+4}\\\\1*(x^2+4x+4) = 2*2\\\\x^2+4x+4 = 4\\\\x^2+4x = 0\\\\x(x+4) = 0\\\\x = 0 \ \text{ or } \ x+4 = 0\\\\x = 0 \ \text{ or } \ x = -4\\\\[/tex]
The two x values, x = 0 and x = -4, make f ' (x) equal to 1/2.
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If x = 0, then y is...
[tex]y = f(x)\\\\f(x) = \frac{x}{x+2}\\\\f(0) = \frac{0}{0+2}\\\\f(0) = 0\\\\[/tex]
Showing that (0,0) is one point we're after.
If x = -4, then,
[tex]y = f(x)\\\\f(x) = \frac{x}{x+2}\\\\f(-4) = \frac{-4}{-4+2}\\\\f(-4) = 2\\\\[/tex]
Making (-4, 2) the other point where the tangent slope is 1/2.
We use the original f(x) function, and not the derivative, when determining the y coordinates of the two points. If you plugged x = 0 or x = -4 into f ' (x), you'd get 1/2. So its good practice, but not what we're after ultimately.
