Find the two x values at which the tangent line to the curve y = (x + 3) / (x + 2) is perpendicular to the line y = x.

1). To be perpendicular to y=x which has a slope of 1 the tangent of the curve must have a slope of - 1

dy/dx = (x+2-x-3) / (x+2) ^2

dy/dx=-1 / (x+2) ^2

dy/dx must equal - 1 to be perpendicular to y=x

-1 / (x+2) ^2=-1

-1 / (x^2+4x+4) = - 1

-1=-x^2-4x-4

x^2+4x+3=0

x^2+x+3x+3=0

x (x+1) + 3 (x+1) = 0

(x+3) (x+1) = 0

x = (-1, - 3)

2). y = x + 3/x + 2

Plots: x from - to - 1.6

Plot: x from - 6 to 1.4

Alternate Form:

y = 1/x + 2 + 1

(x + 2) y = x + 3

Expanded form:

y = x/x + 2 + 3/x + 2

Root: x = - 3

Limit:

3 + x/2 + x = 1