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

the answer is 6

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