Answer
$$
f^{\prime \prime}(x)=8 x^{3}+5 ,\quad f(1)=0 ,\quad f^{\prime}(1)=8
$$
The required function is
$$
f(x)=\frac{2}{5} x^{5}+\frac{5}{2}x^{2}+x-\frac{39}{10}
$$
Work Step by Step
$$
f^{\prime \prime}(x)=8 x^{3}+5 ,\quad f(1)=0 ,\quad f^{\prime}(1)=8
$$
The general anti-derivative of $
f^{\prime \prime}(x)=8 x^{3}+5
$ is
$$
f^{\prime}(x)=2 x^{4}+5x+C
$$
To determine C we use the fact that $f^{\prime}(1)=8 $:
$$
f^{\prime}(1)=2 (1)^{4}+5(1)+C=8
$$
$ \Rightarrow $
$$
7+C=8\quad \Rightarrow \quad C=1,
$$
so
$$
f^{\prime}(x)=2 x^{4}+5x+1
$$
Using the anti-differentiation rules once more, we find that:
$$
f(x)=\frac{2}{5} x^{5}+\frac{5}{2}x^{2}+x+D
$$
To determine D we use the fact that $f(1)=0 $:
$$
f(1)=\frac{2}{5} (1)^{5}+\frac{5}{2}(1)^{2}+(1)+D =0
$$
$ \Rightarrow $
$$
\frac{39}{10}+D=0\quad \Rightarrow \quad D=-\frac{39}{10},
$$
so the required function is
$$
f(x)=\frac{2}{5} x^{5}+\frac{5}{2}x^{2}+x-\frac{39}{10}
$$
