Answer
$f(x) = x^2+1$

Work Step by Step
For $f(0)=1$, the constant term of $f(x)$ must be $1$.
Because $f'(0)=0$, there exists either a relative maximum or minimum at $x=0$.
Because $f'(x)>0$ if $x<0$ and $f'(x)<0$ if $x>0$, the graph is then concave up and $x=0$ is a relative maximum. Thus, we can use the function $f(x) = x^2+1$ which satisfied all of these conditions.