Find the slope of the tangent to the curve:

f(x)=x^3-x^2-x+1

at x= -2

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The slope of the tangent line of `f(x)` at `a` is `f'(a)`

That is, you find the derivaive of `f` and then plug in `a`

So:

`f(x)=x^3-x^2-x+1`

`f'(x)=3x^2-2x-1`

`f'(-2) = 3(-2)^2-2(-2)-1=15`

So the slope of the tangent line of f at -2 is 15.

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