Question

If I have a function f(x) = x^2 * e^-x
and a line L that tangents the graph of the function at x =1. How do I find the function for the line L ?

Answer 1

find the slope of line L by differentiating f(x)
slope = f'(x) at x=1 = f'(1)
can you find f'(1) first ?

Answer 2

once you have slope, you will just need a point on L,
when x=1, find f(1), and the point will be (1, f(1))
try it, if you get stuck , i am here :)

Answer 3

so \[f'(x) = 2xe ^{-x}-x ^{2}e ^{-x}\]

Answer 4

yes, put x=1

Answer 5

I get 0,3678794412 :P

Answer 6

or \[e ^{-1}\]

Answer 7

keep it as 1/e :)

Answer 8

so the point is \[(1,e ^{-1})\]

Answer 9

yes

Answer 10

we have slope, m= 1/e, point = (1,1/e)
just use slope point form
y-y1 = m (x-x1)

Answer 11

could you find the function for line ?

Answer 12

there?

Answer 13

Just looking at ProfRObBob on youtube for the slope-stuff :P I'm norwegian so the english terms are a little different than ours ;)

Answer 14

ohh

Answer 15

y-1/e = 1/e (x-1)
y -1/e = x/e -1/e

y = x/e
or x =ey is the function for line L
hope you got the same :)

Answer 16

Yup, thanks a lot! :)

Answer 17

welcome ^_^