y=(x+3)/(x+2); perpendicular to line y=x?

Question

Find all values of x at which the tangent line to the given curve satisfies the stated property.

I know to use the product rule to find F'(x), but then do I solve the numerator for x??? I just don't understand the question, maybe someone can explain... thanks

I MEANT THE QUOTIENT RULE!!!

Answer 1

for y=x the slope of this line is 1. the perpendicular has the slope -1

So, you have to found the values for which the derivative =1

dy/dx= x+2-x-3/(x+2)^2 =-1/(x+2)^2 =-1
this gives (x+2)^2 =1
or x+2 =1 so x=-1 and x+2 =-1 x=-3

for x=-1 y=-2 and for x=-3 y =0

the 2poitq have the coordinates x=-1 y=2 and x=3 y=0

Answer 2

---
Consider what it means for two lines to be perpendicular.. Their slopes must be negative reciprocals of each other.

example: y = 4x y = -1/4 x 4 is the slope of the first and -1/4 is the slope of the second..
---
So since the slope of the y = x is 1.. the slope of any line perpendicular will be -1.

So find the y' and set it equal to -1.
--
y = (x+3)/(x+2)
y' = (x+2)-(x+3)/[x+2]^2
y' = -1/[x+2]^2

let y' = -1
-1 = -1/[x+2]^2
1 = 1/[x+2]^2
[x+2]^2 = 1
|x+2| = 1
so x can either be -3 or -1