we can apply some rules backwards
first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16
we know that
\(f'(rx^n)=rnx^{n-1}\)
so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3
one is
\(\frac{-2}{3}x^3\)
second part
12x
12x=\(rnx^{n-1}\)
x^1, 1=n-1
n=2
rn=12
2r=12
r=6
\(6x^2\) is the second bit
last part
-16
-16x^0=\(rnx^{n-1}\)
0=n-1
n=1
rn=-16
1r=-16
r=-16
-16x^1
so therfor f(x)=\(\frac{-2}{3}x^3+6x^2-16x\)
