Question

Please help on this problem

Let f(x) = x-x^3 to determine the approximate slope of the line tangent to the graph of f at (a, f (a)).

a = -0.2
slope= ?

&

a = 0.6
slope=?

thanks

Answer 1

I am really lost

Answer 2

0.88 was wrong for some reason

Answer 3

oh ok
well the derivative of f(x) is 1-3x^2

Answer 4

can you show me step by step?

Answer 5

a= 0.6 then the slope is ?

Answer 6

yeah its called the power rule

d/dx x^n = n*x^(n-1)

so for x
x^1 -> 1*x^0 = 1

for x^3
x^3 --> 3*x^2

Answer 7

could you do the same for question part b?

Answer 8

are you sure its not 0.88 ??

Answer 9

i'll plug it in again to be sure

Answer 10

1-3(-.2)^2 = 0.88

Answer 11

I plugged it in and its right for a=-0.2

Answer 12

for part b) its the same derivative function, you are just evaluating the slope at x=0.6

--> 1-3(.6)^2

Answer 13

ok hold on for sec let me work it out before you tell me the anwser lol

Answer 14

I got -0.72 is that right?

Answer 15

nope

Answer 16

awww

Answer 17

multiply before you add

Answer 18

-0.08?

Answer 19

i think you did 1-3 = -2 * .36 ??

multiply 3*.36 first

Answer 20

yes :)

Answer 21

yeahhhh! lol

Answer 22

so the derivitive would be 1-3(x)^2 for this problem only?

Answer 23

yes only for the function x-x^3

Answer 24

ok thanks alot