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- anonymous

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

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- anonymous

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- anonymous

I am really lost

- anonymous

0.88 was wrong for some reason

- dumbcow

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

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- anonymous

can you show me step by step?

- anonymous

a= 0.6 then the slope is ?

- dumbcow

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

- anonymous

could you do the same for question part b?

- dumbcow

are you sure its not 0.88 ??

- anonymous

i'll plug it in again to be sure

- dumbcow

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

- anonymous

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

- dumbcow

for part b) its the same derivative function, you are just evaluating the slope at x=0.6
--> 1-3(.6)^2

- anonymous

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

- anonymous

I got -0.72 is that right?

- dumbcow

nope

- anonymous

awww

- dumbcow

multiply before you add

- anonymous

-0.08?

- dumbcow

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

- dumbcow

yes :)

- anonymous

yeahhhh! lol

- anonymous

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

- dumbcow

yes only for the function x-x^3

- anonymous

ok thanks alot

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