Answer
$$
f^{\prime \prime}(x)=4+6 x+24 x^{2} ,\quad f(0)=3 ,\quad f(1)=10
$$
The required function is
$$
f(x)=2x^{2} + x^{3}+2x^{4}+2x+3
$$
Work Step by Step
$$
f^{\prime \prime}(x)=4+6 x+24 x^{2} ,\quad f(0)=3 ,\quad f(1)=10
$$
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
$$
f(x)=2x^{2} + x^{3}+2x^{4}+2x+3
$$
