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