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