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 anti-derivatives 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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