Answer
$$
f^{\prime \prime}(x)=5 x^{3}+6x^{2}+2 ,\quad f(0)=3 ,\quad f(1)=-2
$$
The required function is
$$
f(x)=\frac{1}{4} x^{5}+\frac{1}{2}x^{4}+x^{2}+Cx-\frac{27}{4}
$$
Work Step by Step
$$
f^{\prime \prime}(x)=5 x^{3}+6x^{2}+2 ,\quad f(0)=3 ,\quad f(1)=-2
$$
The general anti-derivative of $
f^{\prime \prime}(x)=5 x^{3}+6x^{2}+2
$ is
$$
f^{\prime}(x)=\frac{5}{4} x^{4}+2x^{3}+2x+C
$$
Using the anti-differentiation rules once more, we find that:
$$
f(x)=\frac{1}{4} x^{5}+\frac{1}{2}x^{4}+x^{2}+Cx+D
$$
To determine $ C, D$ we use the fact that $f(0)=3, f(1)=-2$:
$$
f(0)=\frac{1}{4} (0)^{5}+\frac{1}{2}(0)^{4}+(0)^{2}+C(0)+D=3
$$
$ \Rightarrow $
$$
0+D=3 \quad \Rightarrow \quad D=3,
$$
and
$$
f(1)=\frac{1}{4} (1)^{5}+\frac{1}{2}(1)^{4}+(1)^{2}+C(1)+3=-2
$$
$ \Rightarrow $
$$
\frac{19}{4}+C=-2 \quad \Rightarrow \quad C=-\frac{27}{4},
$$
so the required function is
$$
f(x)=\frac{1}{4} x^{5}+\frac{1}{2}x^{4}+x^{2}+Cx-\frac{27}{4}
$$
