Answer:
[tex]y = \frac{1}{2}x+\frac{7}{2}[/tex]
[tex]y = \frac{1}{2}x-\frac{1}{2}[/tex]
Step-by-step explanation:
The first step is obtain the derivative of the function y:
[tex]\frac{dy}{dx}=\frac{(x+1)*(1)-(x-1)*1}{(x+1)^{2} }=\frac{2}{(x+1)^{2} }[/tex]
(Remember that the slope of the tangent line to any curve is given by its derivative).
Secondly, in order to obtain the equations of the tangent lines parallel to x - 2y = 5 we need to obtain the slope of this line, remember that the line equation is given by:
[tex]y = mx + b[/tex]
m : slope
b: y-intercept
Then, the slope of the line
[tex]y = \frac{1}{2}x-\frac{5}{2}[/tex]
is m = 1/2.
Then, the derivative of the function y must be 1/2:
[tex]y= \frac{2}{(x+1)^{2}}=\frac{1}{2}\\(x+1)^{2}=4\\x^{2}+2x-3=0\\ (x+3)(x-1)=0\\x = -3\\x=1[/tex]
When x = -3, y =(-3-1)/(-3+1)= 2, then one of the tangent lines must pass through the point (-3,2).
When x = 1, y = (1-1)(1+1)= 0, then the other tangent line must pass through the point (1, 0).
Finally, the point-slope equation of a line is given by:
[tex]y- y_{1} = m(x - x_{1})[/tex]
Substituting our previous results we have the equations of the tangent lines:
[tex]y - 2 = \frac{1}{2}(x - (-3))\\ y = \frac{1}{2}x+\frac{7}{2}[/tex]
[tex]y - 0 = \frac{1}{2}(x- 1) \\y = \frac{1}{2}x-\frac{1}{2}[/tex]
