Let $f\left(x\right)=x^4-4x^3+6x^2-4x+1$then

$f\text{'}\left(x\right)=4(x^3-3x^2+3x-1)=4{(x-1)}^3>0$ for x > 1

$f$increases is $[1,\propto ),f\text{'}\left(x\right)<0$ for $x<1$

$f$has a minimum at x = 1.