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Find all points on the graph of f(x)=-x^3+3x^2-2 where there is a horizontal tangent line.
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The tangent line equation is y=f'(x)(x-x0)+f(x0) but i don understand the question.
f'(x)=-3x^2+6x, that's all I got. You suppose to find all the points. A. (0, -2), (2, 2) B. (0, -2) C. (1, 0), (0, -2) D. (2, 2) E. None of these
Horizontal tangents occur where the derivative / slope of the function is 0. We simply need to take the derivative of our function to find the equation of slope at any point. Set the derivative equal to 0 and solve for x for the points where the slope is horizontal / 0.
I got it. Thanks.
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