Question

Let f(x)=x^3-x-1
A. Find the equation of the tangent line to the graph of the function at x=0

Answer 1

find f'(x) in terms of x and put x = 0
this will give you the slope of the tangent line

Answer 2

3x^2-1

Answer 3

so when x = 0 slope is 3(0) - 1 = -1

Answer 4

that is the answer -1
I would write the answer as follow so when x=0 slope is 3(0)-1=-1

Answer 5

no -1 is the slope of the line

use the general form of a straight line

y-y1 = m(x-x1)

m = slope = -1
and (x1,y1) is the coordinates of the point of contact of the tangent line
we know the x1 = 0
so we need to find the value of y1

how would we do that?

Answer 6

I am not sure really

Answer 7

just put x = 0 in x^3 -x-1

Answer 8

6x-1

Answer 9

?? hoe did you get 6x??

Answer 10

3x^2

Answer 11

0^3 - 0 -1 = ?

Answer 12

4

Answer 13

how did you get 4?

0^3 = 0

Answer 14

see I really don't know what I am doing

Answer 15

y=0

Answer 16

y = 0 - 0 -1 = -1

Answer 17

now insert m = -1 , x1 = 0 and y1 = -1 into the equation to get the required equation

Answer 18

y - y1 = m(x - x1)

y - (-1) = -1(x - 0)

Answer 19

now simplify

Answer 20

this is what i get out of this y=-1=-1 i know this isn't right