How do I go about finding all the values of x where the tangent line to y = f(x) is horizontal? f(x) = x^3(2x^2 + x - 3)^2 I assume I would set f(x) and f '(x) equal to 0 and solve for x?
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I can't help you because I don't know derivatives however, as I recall from previous posts, you need to find the value of x when the f '(x)=0 finding when f(x)=0 does nothing.
Last edited by Quick; Aug 26th 2006 at 06:27 AM.
Quick is right the question is asking to find all the points where f '(x) = 0 Use product rule (with chain rule needed for differentiating v) Factorising gives gives gives gives Gives and
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