Answer
$f(x)=-\sin x+\frac{3}{2}x^2 +3x +1$
Work Step by Step
$$f'''(x)=\cos x$$. $$\text{then} \ f''(x)= \sin x +C$$.
$$3= \sin0 +C \Rightarrow C=3.$$
$$f'(x)=-\cos x+3x+C$$
$$2=-\cos0+3\cdot0 +C \Rightarrow C=3$$
$$f(x)=-\sin x +\frac{3}{2}x^2+3x+C$$
$$1=-\sin0+\frac{3}{2}\cdot0^2+3\cdot0+C \Rightarrow 1=C$$
Thus $f(x)=-\sin x+\frac{3}{2}x^2 +3x +1$.
