Question 220705
For what values of x does the graph of f(x) = 2x^3 - 3x^2 - 6x + 51 have a horizontal tangent?
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Take the derivative; set it to zero; solve for "x".
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f'(x) = 6x^2-6x-6 = 0
x^2-x-1 = 0
x = [1 +- sqrt(1-4*1*-1)]/2
x = [1 +- sqrt(5)]/2
x = (1+(sqrt(5))/2 or (1-(sqrt(5))/2
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Let's see what it looks like:
{{{graph(400,300,-10,10,-60,60,2x^3-3x^2-6x+51)}}}
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Cheers,
Stan H.