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I am trying to find $f(x)$ if I know that $f''(x) = -2+12x-12x^2, \quad \; f(0)=4,\ f'(0)=12 $

First I found the first derivative $ f'(x)= -2x+6x^2-4x^3+C$

and then I found the function, which is: $f(x)=-x^2+2x^3-x^4+Cx+D$

Now I am lost as to what to do with those values they gave me $ f(0)=4,\ f'(0)=12 $

Where do i proceed from here?

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    first, in the $f'$ equation, plug in the point $(0,12)$ and solve for the constant $C$. Then do the same for $D$ in the $f(x)$ equation and you're done!2012-07-11

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After finding the value of $f'(x)$ with the unknown constant $C$, use the fact that $f'(0)=12$ to determine the value of $C$. That is, since $f'(x) = -2x + 6x^2 - 4x^3 + C\quad\text{and}\quad f'(0)=12$ that means that $12 = f'(0) = -2(0) + 6(0)^2 - 4(0)^3 + C.$ This should tell you the value of $C$.

Then find $f(x)$, which will give you another unknown constant $D$. Use the fact that $f(0)=4$ to figure out the value of $D$.

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    Give it a spin....2012-07-11