Find equations of the tangent lines to the graph of f(x)= x/x+2 that are parallel to the line x-2y=1

Question

Can someone please help. Thank you.

Answer 1

I will assume that the correct expression for f(x) is x/(x+2).   If we use the quotient rule 

we get f′(x)=(1(x+2)-x·1)/(x+2)²=2/(x+2)^2.  We are asked to find points on the graph of f where the tangent lines are parallel to the graph of the solutions of x-2y=1.  The second graph is a line with slope 1/2.  The equation f′(x)=1/2 or 2/(x+2)²=1/2 is equivalent to (x+2)²=4.  By thinking we can see that the solutions of this equation are ˜4 and 0.  This gives us the first coordinates of points on the graph of f where the tangent lines have the required slope of 1/2.  f(-4)=2 and f(0)=0, so equations for the tangent lines are (y-2)=1/2·(x+4) and (y-0)=1/2·(x-0).  

Answer 2

 

Hi Bob,

 

Let's find out lines tangent to f(x) = x/(x+2) that are also parallel to x - 2y = 1.

 

f(x) can be re-written as follows

 

     f(x) = x/(x+2)

        

            = 1 - 2/(x+2)

 

And taking the first order derivative of f(x), we get

 

     f'(x) = d/dx{1 - 2/(x+2)}

 

            = -2·d/dx{1/(x+2)}

 

            = -2·d/du(1/u)·du/dx     (where, u ≡ x + 2)

 

            = 2/(x+2)²

 

As the lines to be found are parallel to x - 2y = 1,

f'(x) needs to equal 1/2 at those tangential contact points.

 

     f'(x) = 2/(x+2)²

 

            = 1/2

 

Solving the equation above,

we get x = -4 or x = 0 which are two tangential contact points between f(x) and the lines to be found.

 

Both lines have the slope of 1/2, and

one passes through the point (-4, 2) and the other passes through (0, 0).

 

The final answer to the problem includes

 

     y = x/2 + 4, and y = x/2