# calculus

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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 -

you know that
y' = 6x + C
and that y'(0) = 0, so
0 = C

y = 2x^3 + 1

• calculus -

So y'(0) = 0 because there is a horizontal tangent at x = 0?

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