[tex]f(x)=(x-1)(x-4)^2[/tex]
[tex]f'(x)=(x-4)^2+2(x-1)(x-4)=(x-4)(x-4+2x-2)=(x-4)(3x-6)=3(x-4)(x-2)[/tex]
The first derivative is zero when [tex]x=2[/tex] and [tex]x=4[/tex], so you need to only check the sign of the derivative in the intervals [tex](-\infty,2)[/tex], [tex](2,4)[/tex], and [tex](4,\infty)[/tex].
Since [tex]f'(x)[/tex] describes a convex function (a parabola that opens upward), and you know the location of the roots, you know that [tex]f'(x)<0[/tex] on the interval [tex](2,4)[/tex], and so the function is decreasing on this interval only.
