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anonymous
 5 years ago
Find the two horizontal tangents on the graph of f(x)=2x^3+12x^272x+16
anonymous
 5 years ago
Find the two horizontal tangents on the graph of f(x)=2x^3+12x^272x+16

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find the derivative of this function to be: f'(x) = 6x^2 + 24x  72 Horizontal tangents occur when the slope of the tangent line is 0 (horizontal lines have a slope of 0). So, substitute 0 for f'(x) and solve for x. 0 = 6x^2 + 24x  72 0 = x^2 + 4x  12 0 = (x + 6)(x  2) x = 2, 6 One horizontal tangent line intersects the function at x = 2, and the other at x = 6. Find there corresponding f(x) values to be f(2) = 64 and f(6) = 448. So, the horizontal tangents are y = 64 and y = 448
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