Posted by asdf on Thursday, November 29, 2012 at 5:04pm.
Find a function f such that the curve y = f(x) satisfies y'' = 12x, passes through the point (0,1), and has a horizontal tangent there.
I can use the antiderivatives to get
y = 2x^3 + Cx + D
but I don't know how to get C even though I know D = 1.

calculus  Steve, Thursday, November 29, 2012 at 5:08pm
you know that
y' = 6x + C
and that y'(0) = 0, so
0 = C
y = 2x^3 + 1

calculus  asdf, Thursday, November 29, 2012 at 5:56pm
So y'(0) = 0 because there is a horizontal tangent at x = 0?
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