Answer
$$
f^{\prime \prime}(x)=1-6x +48 x^{2} ,\quad f(0)=1 ,\quad f^{\prime}(0)=2
$$
The required function is
$$
f(x)=\frac{1}{2}x^{2}-x^{3} +4x^{4}+2x+1.
$$
Work Step by Step
$$
f^{\prime \prime}(x)=1-6x +48 x^{2} ,\quad f(0)=1 ,\quad f^{\prime}(0)=2
$$
The general anti-derivative of $
f^{\prime \prime}(x)=1-6x +48 x^{2}
$ is
$$
f^{\prime}(x)=x-3x^{2} +16 x^{3}+C
$$
To determine C we use the fact that $f^{\prime}(0)=2 $:
$$
f^{\prime}(0)=(0)-3(0)^{2} +16 (0)^{3}+C =2
$$
$ \Rightarrow $
$$
0+C=2\quad \Rightarrow \quad C=2,
$$
so
$$
f^{\prime}(x)=x-3x^{2} +16 x^{3}+2
$$
Using the anti-differentiation rules once more, we find that:
$$
f(x)=\frac{1}{2}x^{2}-x^{3} +4x^{4}+2x+D
$$
To determine D we use the fact that $f(0)=1 $:
$$
f(0)=\frac{1}{2}(0)^{2}-(0)^{3} +4(0)^{4}+2(0)+D=1
$$
$ \Rightarrow $
$$
0+D=1\quad \Rightarrow \quad D=1,
$$
so the required function is
$$
f(x)=\frac{1}{2}x^{2}-x^{3} +4x^{4}+2x+1.
$$
