f(x)=|x−3|     x≥1x24 − 3x2+134   x<1     is

limx→1−f(x) = limx→1−(x24−3x2+134)= 14−32+134=2 limx→1+f(x)=limx→1+|x−3|=2(continuous at x=1) f1(1−)=limx→1−x24−3x2+134−2x−1=limx→1−(x−1)(x−5)4(x−1)=−1 f1(1+)=limx→1+|x−3|−2x−1=limx→1+−(x−1)x−1=−1 (Differentiable at x = 1|x−3|  is continuous at x = 3 but not differentiable at x = 3