Question

Find the equations of the tangents to the given curves for the given values of x.

(a) y=e^x, where x=-1
(b) y=2x-e^-x, where x=0

Answer 1

Tell me if I'm wrong but are you suppose to find the derivatives of each and and find the values given x?

Answer 2

No... you're supposed to find the equation of the tangents... eg. the answer of the first is ey=x+2 (don't know how to get to the answer)

Answer 3

first the derivative of the first one is e^x

Answer 4

The slope at the point x = -1, is e^-1

Answer 5

going back to the original equation, the (x,y) at x = -1 is (-1, e^-1)

Answer 6

using the point slope formula. (y - y1) = m(x - x1) we will solve the tangent line equation

Answer 7

(y- e^-1) = e^-1(x + 1)
multiply both sides by e
ey - 1 = x + 1
add to both sides
ey = x + 2

Answer 8

Thats your answer. You can go ahead and put it in the form "y = mx + b" if you want to. If you do so it will be y = x/e + 2/e

Answer 9

I get everything up to
(y- e^-1) = e^-1(x + 1)
multiply both sides by e
ey - 1 = x + 1
add to both sides
ey = x + 2

Answer 10

what i am doing here is that i am using the point slope formular to solve for the equation of the tangent line. Slope is m ==> e^-1; y1 = e^-1; x1 = -1

Answer 11

Yes, I understand... but why is this this?
e^-1(x+1)
That's what I don't understand..

Answer 12

It is the m(x - x1) ==> (e^-1)(x - (-1) )

Answer 13

How did you arrive at that solution?

Answer 14

for the m(the slope, at the point where x = -1) i used the derivative of e^x, and plugged in -1.

Answer 15

Thanks, I understand that. And for the m1?

Answer 16

there is no m1

Answer 17

(x - (-1) I mean M(x1)

Answer 18

How did you get (x - (-1) ?

Answer 19

(x - x1) my (x1 = -1)

Answer 20

Um... I still don't fully understand?

Answer 21

Are u familiar with the point slope formula

Answer 22

Yes, \[{x2-x1 \over y2-y1}=m\]

Answer 23

I mean the other way

Answer 24

y is on top

Answer 25

that's the slope formula. But the point slope can be derived from it.

Answer 26

Yes, by y2-y1=m(x2-x1)?

Answer 27

Remember to get the equation of a line, using the point slope formula we need a point and a slope. In this case since we are looking for the tangent line at x = -1. Our point would be (-1, e^(-1)) and our slope at x = -1 is e^-1

Answer 28

Ooh! Ok. I understand now! :)

Answer 29

I am glad that helped

Answer 30

from this, (y- e^-1) = e^-1(x + 1)
How do you get the answer? I'm sorry for bothering you!