f(x) = x-x^3 a = -0.2 slope =??

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f(x) = x-x^3 a = -0.2 slope =??

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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could you clarify the question? what does a refer to?
tangents
Let f(x) = x-x^3. determine the approximate slope of the line tangent of f at (a, f (a)).

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a = -0.2 slope=?
I think derivitives
ok thank you the slope of f(x) can be determined by taking derivative f'(x) =1 - 3x^2 now plug in x = -0.2 f'(-0.2) = 1 - 3(-.2)^2 = 0.88
so 0.88 would be the slope?
yes f'(x) is same thing as slope at point (x, f(x))
thank you
do you know how to take the derivative of polynomials?
d/dx x^n = n*x^(n-1)

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