Calculus 8th Edition

$$f^{\prime \prime}(x)=4+6 x+24 x^{2} ,\quad f(0)=3 ,\quad f(1)=10$$ The required function is $$f(x)=2x^{2} + x^{3}+2x^{4}+2x+3$$
$$f^{\prime \prime}(x)=4+6 x+24 x^{2} ,\quad f(0)=3 ,\quad f(1)=10$$ The general anti-derivative of $f^{\prime \prime}(x)=4+6 x+24 x^{2}$ is $$f^{\prime}(x)=4x +3 x^{2}+8x^{3}+C$$ Using the anti-differentiation rules once more, we find that: $$f(x)=2x^{2} + x^{3}+2x^{4}+Cx+D$$ To determine D, C we use the fact that $f(0)=3, f(1)=10$: $$f((0))=2(0)^{2} + (0)^{3}+2(0)^{4}+C(0)+D=3$$ $\Rightarrow$ $$0+D=3\quad \Rightarrow \quad D=3,$$ so $$f(x)=2x^{2} + x^{3}+2x^{4}+Cx+3$$ and $$f(1)=2(1)^{2} + (1)^{3}+2(1)^{4}+C(1)+3=10$$ $\Rightarrow$ $$8+C=10 \quad \Rightarrow \quad C=2,$$ so the required function is $$f(x)=2x^{2} + x^{3}+2x^{4}+2x+3$$