Answer
$$
f^{\prime \prime}(x)=20 x^{3}+12x^{2}+4 ,\quad f(0)=8 ,\quad f(1)=5
$$
The required function is
$$
f(x)=x^{5}+x^{4}+2x^{2}-7x+8
$$
Work Step by Step
$$
f^{\prime \prime}(x)=20 x^{3}+12x^{2}+4 ,\quad f(0)=8 ,\quad f(1)=5
$$
The general anti-derivative of $
f^{\prime \prime}(x)=20 x^{3}+12x^{2}+4
$ is
$$
f^{\prime}(x)=5x^{4}+4x^{3}+4x+C
$$
Using the anti-differentiation rules once more, we find that:
$$
f(x)=x^{5}+x^{4}+2x^{2}+Cx+D
$$
To determine $ C, D$ we use the fact that $f(0)=8, f(1)=5$:
$$
f(0)=(0)^{5}+(0)^{4}+2(0)^{2}+C(0)+D=8
$$
$ \Rightarrow $
$$
0+D=8 \quad \Rightarrow \quad D=8,
$$
and
$$
f(1)=(1)^{5}+(1)^{4}+2(1)^{2}+C(1)+8=5
$$
$ \Rightarrow $
$$
12+C=5 \quad \Rightarrow \quad C=-7,
$$
so the required function is
$$
f(x)=x^{5}+x^{4}+2x^{2}-7x+8
$$
