Answer
$$f(x) =2 x^{3}-2 x^2+3x+1 $$
Work Step by Step
Given $$f^{\prime}(x)=6 x^{2}-4x+3$$
Since
\begin{align*}
f(x)&=\int f'(x)dx\\
&= \int [6 x^{2}-4x+3]dx\\
&= 2 x^{3}-2 x^2+3x+C
\end{align*}
Since $f(0)= 1$, then $C= 1$ and
$$f(x) =2 x^{3}-2 x^2+3x+1 $$
