## Calculus 8th Edition

$$f^{\prime \prime}(x)=-2+12 x-12 x^{2} ,\quad f(0)=4 ,\quad f^{\prime}(0)=12$$ The required function is $$f(x)=-x^{2}+2 x^{3}-x^{4}+12 x+4$$
$$f^{\prime \prime}(x)=-2+12 x-12 x^{2} ,\quad f(0)=4 ,\quad f^{\prime}(0)=12$$ The general anti-derivative of $f^{\prime \prime}(x)=-2+12 x-12 x^{2}$ is $$f^{\prime}(x)=-2 x+6 x^{2}-4 x^{3}+C$$ To determine C we use the fact that $f^{\prime}(0)=12$: $$f^{\prime}(0)=-2 (0)+6(0)^{2}-4 (0)^{3}+C =12$$ $\Rightarrow$ $$0+C=12\quad \Rightarrow \quad C=12,$$ so $$f^{\prime}(x)=-2 x+6 x^{2}-4 x^{3}+12$$ Using the anti-differentiation rules once more, we find that: $$f(x)=-x^{2}+2 x^{3}-x^{4}+12 x+D$$ To determine D we use the fact that $f(0)=4$: $$f(0)=-(0)^{2}+2 (0)^{3}-(0)^{4}+12 (0)+D =4$$ $\Rightarrow$ $$0+D=4\quad \Rightarrow \quad D=4,$$ so the required function is $$f(x)=-x^{2}+2 x^{3}-x^{4}+12 x+4$$