Answer
$$f(t) =\frac{1}{2}t^2- \frac{1}{3}t^3-t+2 $$
Work Step by Step
Given$$
f^{\prime \prime}(t)=1-2 t, f(0)=2, \text { and } f^{\prime}(0)=-1
$$
Then
\begin{align*}
f'(t) &= \int (1-2t)dt\\
&= t-t^2+c
\end{align*}
At $t= 0$, $f'(0)=-1,$ then $c= -1$, hence
$$f'(t)= t-t^2-1 $$
\begin{align*}
f(t) &= \int ( t-t^2-1 )dt\\
&= \frac{1}{2}t^2- \frac{1}{3}t^3-t+c_1
\end{align*}
At $t=0$ $f=2$, then $c_1= c_1=2 $, hence
$$f(t) =\frac{1}{2}t^2- \frac{1}{3}t^3-t+2 $$
