Explanation:

f(x)=x⁴−8x³+22x²−24x+4
f′(x)=4x³−24x²+44x−24
and f”(x)=12x²−48x+44
f’(x)=0
=> 4x³−24x²+44x−24=0  or (x−1)(x−2)(x−3)=0
The critical points are x=1,2,3
Putting these points in f’’(x) we get f(1)=+ve, f(2)=-ve, f(3)=+ve
So, at x=1 and 3 there is local minima and x=2 there is local maxima. This means, the graph would be increasing beyond x=1 and x=3