Question

let y=(e^x+e^x)/2
c. Find an equation of the line tangent to the curve at x = 1.
d. Find an equation of the line normal to the curve at x = 1.
e. Find any points where the tangent line is horizontal.

Answer 1

y=(e^x+e^x)/2=e^x
For the tangent line you need
d/dx e^x=?

Answer 2

finding dy/dx and d^2y/dx^2 was the first part in case you need it.
dy/dx=e^x
d^2y/dx^2=e^x

Answer 3

Correct.
What do you need to find an equation for the tangent line at x=1?? (2 data)

Answer 4

I don't know

Answer 5

A point on the line and the slope of the line.
Can you find these?

Answer 6

Are you trying at least?

Answer 7

well its just a line right?

Answer 8

oh wait its not a line its a curve

Answer 9

Find an equation of the line tangent to the curve at x = 1.

This is what we try to find.
For any straight line you can easily get an equation if you know a point on the line and the slope of the line.
This is a special line as it is tangent to the curve.

Answer 10

at x=1, the curve is at around 2.5?

Answer 11

If you do not know how e^x looks like, just type into Google: graph e^x

Now back to this problem.
So e^x and the tangent to the curve at x=1 have some special properties.
They have the same slope at that point. (x=1)
And they take the same value there

Answer 12

at x=1 the value is e^1=e
You can leave it like that ( I prefer it that way), or
\[e \approx 2.71828\]

Answer 13

So we have found a point!! At x=1 y=e
(1,e)
Can you find the slope now?

Answer 14

the slope would be e too right?

Answer 15

Yes!

Answer 16

So what is the equation of the tangent line?

Answer 17

\[y-y _{1}=m(x-x _{1})\]
This is the formula you need here

Answer 18

ok would it be ex?

Answer 19

Yes, y=ex

Answer 20

so that's the answer to the first part? y=ex?

Answer 21

OK can you see why the solution of d, is
y=-ex ?

Answer 22

not really

Answer 23

wait I might be wrong here :)

Answer 24

Lets do this step by step, first do you know what a normal line is?

Answer 25

yes

Answer 26

OK so we know that it goes through the same point (1, e)
But it wont have slope m=e, but ?

Answer 27

In other words, Lamp, a tangent line and a normal line to a curve at a certain point (1,e) have slopes that are related to one another. Do you recall what that relationship is?

Answer 28

no but I think the slope is e^x?

Answer 29

Lamp, Andras has already helped you to find the slope of the tangent line to the graph of y=e^x. It's e^x evaluated at x=1. Now you're asked to find the slope of the normal line to the graph at the same point. Again, I ask you how the slopes of perpendicular lines are related.

Answer 30

No, If you need to find the slope of a line perpendicular to another, you need to take the negative reciprocal.
So in this case the m will change to -1/e

Answer 31

If I were in your shoes and I didn't remember, I'd do a quick Internet search for "slopes of perpendicular lines." Andras is again right on track; the key words here are "negative reciprocal."

Answer 32

@lamp12345 you are very attractive just sayin

Answer 33

ok with that slope I got -x/e+e+1/e

Answer 34

Nice, you know maths :D

Answer 35

yay! is the last part similar the the first 2?

Answer 36

Last question is a bit different, any guess?

Answer 37

First try to find the slope of the line that is horizontal

Answer 38

the normal line is horizontal, so would I find the slope of that?

Answer 39

Or, in other words, what's the slope of a horizontal line?

Answer 40

ok I got a slope of 1/3

Answer 41

I mean -1/3 @Andras

Answer 42

No that is wrong


Is this horizontal?

Answer 43

no

Answer 44

m has to be 0, if you want a horizontal line.

Answer 45

ok makes sense

Answer 46

Lamp, please look up the words "horizontal" and "vertical". It's essential that you know the meaning of "horizontal" before y ou try to solve a problem involving a horizontal line.

Could you answer these questions:
What is the slope of a horizontal line?
What is the slope, if any, of a vertical line?

Answer 47

tangent line is horizontal

This tells us that
d/dx =0
In this case
e^x=0
Can you solve that?

Answer 48

x would be 0 right?

Answer 49

Lamp I have to go now so I finish this quick.
The above equation
e^x=0 has no solutions!!!
This means that there is no horizontal tangent line to e^x

If you look at the graph of e^x you can believe this easily.

Answer 50

any number to the power of 0=1
like e^0=1

Answer 51

ok so there would be no solution.

Answer 52

yep

Answer 53

now what?

Answer 54

@Andras does that mean there's no points and it doesn't exist?

Answer 55

No. Andras was correct in stating that YOUR GRAPH HAS NO HORIZONTAL TANGENT LINE. Why? Because the equation e^x=0 has no solution.