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anonymous

  • 5 years ago

Find the two horizontal tangents on the graph of f(x)=2x^3+12x^2-72x+16

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  1. anonymous
    • 5 years ago
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    find 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

  2. anonymous
    • 5 years ago
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    o great, thanks!!

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