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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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Plug in 0. And I assume the first $-2x$ is supposed to be $-2$? – Mike Jul 11 '12 at 1:05
Nope it should be $-2x$ since im finding the antiderivative right? – soniccool Jul 11 '12 at 1:07
So i should plug in 0 for the second and 0 for the first and get those answers back? – soniccool Jul 11 '12 at 1:07
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! – Robert Mastragostino Jul 11 '12 at 1:09

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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So i get the calue C = 12. Then i plug in 12 into C into the f(x) and get D right? – soniccool Jul 11 '12 at 1:24
Give it a spin.... – ncmathsadist Jul 11 '12 at 1:37

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