Answer
$f_{1},f_{2},f_{3},f_{4}$ are linearly independent.
Work Step by Step
$af_{1}(t)+bf_{2}(t)+cf_{3}(t)+df_{4}=\;a(2t-3)\;+b(t^3+1)\;+c(2t^2-t)\;+d(t^2+t+1)=\;0\\\\$
$(-3a+b+d)+\;(2a-c+d)t+(2c+d)t^2+bt^3=0\\\\$
By the polynomial equality theorem:
$\left\{\begin{matrix}
-3a+b+d=0 \\
2a-c+d=0 \\
2c+d=0 \\
b=0
\end{matrix}\right.
\rightarrow \;\;\;\;a=0\;\;\;b=0\;\;\;c=0\;\;\;d=0\\\\$
so, $f_{1},f_{2},f_{3},f_{4}$ are linearly independent.
