Question

simple but confused me :(

find the slope of f(x) at the point where it crosses the x-axis

f(x)=1-e^x
------------
I'm a little bit confused,
now the function or graph is going to cross the x-axis at the points of its roots, i.e when f(x)=0

is that right :d

so
1-e^x=0
e^x=1
x=ln1
x=0

so the root is zero, and the graph is going to cross the x-axis when x=0 ???


the slope at that point is going to be the derivative of the function at the same point,
f'=e^x

so the slope :
=e^0 = 1

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is my work OK, or there is a big misunderstanding :(

Answer 1

here we go, the king is around :)
how r u man :)

Answer 2

who r u referring to? i hope not me because this is all a guess.

Answer 3

kidding!!

Answer 4

check your derivative... that's not right.

Answer 5

is it -e^x ?

Answer 6

yes. so the slope is -1

Answer 7

what about the equation of the tangent line to the curve at this point !

Answer 8

what are the x, y coordinates of the point in question?

Answer 9

the question says when the graph cross the x-axis, there is no x,y coordinates

Answer 10

every point on the graph has an x,y coordinate

Answer 11

this is my confusing, how to get the x, y of the point, I'm suppose to get the coordinates by finding the point where the graph cross the x-axis

Answer 12

I'm really confused :(

Answer 13

ok... the point in question is the point where the graph crosses the x-axis.... you already have the x-coordinate of where that happens... what is it?

Answer 14

y=0 x=0

Answer 15

yes.....

Answer 16

we have the slope -1

Answer 17

so, you have a point (0, 0), and a slope at that point m=-1... how d u use this info to come up with the equation of a line?

Answer 18

is it y=mx+b so 0=-1(0)+b lool

Answer 19

horrible work I now :D

Answer 20

do you know point-slope form for the equation of a line?
y -y0 = m*(x - x0)

Answer 21

you don't have to use y=mx + b..

Answer 22

y -y0 = m*(x - x0) where (x0, y0) is the point you know and m is the slope you know...

Answer 23

y-0=-1(x-0)

Answer 24

ok... you can leave it like that. but do you think you can simplify that a bit?

Answer 25

yes I'll try now

Answer 26

y-0=-x-0

Answer 27

how about y = -x

Answer 28

good :)

Answer 29

other than your derivative being miscalculated your work looks fine...
good work man...

Answer 30

you r just great man, the way you teach is amazing, wish all the best in ur life :)

Answer 31

and I hope me guessing how to do this by the seat of my pants yielded correct results.

Answer 32

kidding!!!

Answer 33

lool

Answer 34

thanx for ur time

Answer 35

np & yw...

Answer 36

\[f(x) =1-e^x\\
f(0) = 0\\
\text { You need to find } f"(0).\\
f'(x)= - e^x\\
f'(0) = -1
\]

Answer 37

You need to find f'(0)
There was a misprint above.