The function f(x)=|x−3|,x≥1x24−3x2+134,x<1

f'1−    =limx→1−f(x)−f(1)x−1                    =limx→1x24−3x2+134−2x−1                   = limx→1x2−6x+54(x−1)                  =limx→1(x−5)(x−1)4(x−1)=limx→1(x−5)4=−1f'1+=limx→1+f(x)−f(1)x−1limx→13−x−2x−1=−1∴f'(1)=−1∴f(x) is derivable and hence continuous also at x=1