Question

find the values of x at which the tangent line to the graph is parallel to the line y=x.
y= (x^2+1)/(x+1)

Answer 1

wait would the derivative be 1? meaning it would be a line meaning no tan line?

Answer 2

y-x has a slope of 1
so if you differentiate y= (x^2+1)/(x+1)
and equate it to and then solve for x
you will get the required values of x

Answer 3

* equate it to 1

Answer 4

well for the derivative I got the answer: x^2+2x+1/(x+1)^2 which would be 1 because it is the same top and bottom?

Answer 5

the answer in the back says none, i'm just tryind to figure out why.

Answer 6

i'll just check it out again

Answer 7

ok so if you set the derivative to 1 then both sides would be equal to each other. leaving nothing. is that how that would be done?

Answer 8

whats the matter with me today - sorry - the derivative y' = 1 is a line of slope 1 which is a line parallel to y=x. two parallel liones never intersect so there are no solutions

Answer 9

how wud you do a problem like this when it asks for it to be perpendicular? its y=(x+3)/(x+2)

Answer 10

perpendicular to y=x?

Answer 11

yes. i got the derivative as -1/(x+2)^2

Answer 12

yes derivative is correct

Answer 13

so what would I be setting it equal to for a perpendicular line? would be be the opposite reciprical?

Answer 14

the derivative gives the slope of the tangent to the curve

Answer 15

so y = x always has the slope of 1? so would you set it to -1?

Answer 16

Sorry, I'm so confused.

Answer 17

the derivative -1/(x+2)^2 gives us the slope of all points x on the curve (x+3)/(x+ 2)
eg at x = 1 the slope is -1 /(1+2)^2 = -1/8. we are required to find values of x which will give a line perpendicular to y = x , right?

Answer 18

yes

Answer 19

ok the slope of a line perpendicular to a line of slope m = -1/m.
in case of y=s the slope is -1
so -1/(x+2)^2 = -1
-(x+2)^2 = -1
(x+2)^2 = 1
x^2 + 4x + 3 = 0
solve this to get your answers

Answer 20

x = -1 , -3

Answer 21

ok?

Answer 22

yes that is what I got. Thank you for helping me