If $f(x)$ is a monic quartic polynomial such that $f(-1)=-1$, $f(2)=-4$, $f(-3)=-9$, and $f(4)=-16$, find $f(1)$.
We can use the finite difference method:
f(-1) f(2) f(-3) f(4) f(1)
-1 -4 -9 -16 -1
-3 -5 -7 15
-2 -2 22
0 24
24
Therefore f(1) = -1.
Can you please elaborate on your solution?
how does 24 made f(-1)=1?
I get the following
