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Find the points on the graph of f(x)= 12(x + 8) − (x + 8)^3 where the tangent line is horizontal. (x, f(x))= (smaller x-value) (x, f(x))=(larger x-value)
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so find the derivative, let it equal zero and solve for x. \[f ' (x) = 12 -3(x+8)^2\] so \[f '(x) = -3x^2 -48x - 180\] then solve \[-3(x^2 + 16x + 60) = 0\]
so what is it 6, 10?
so (x+6)(x+10) the solutions are x = -6, -10 so the tangents are horizontal at x = -6 and x = -10
dang, for some reason the stupid homework says its wrong :/
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