Answer :

Consider, f(x) = |x-1| + |x-2|


Let’s discuss the continuity of f(x).


We have, f(x) = |x-1| + |x-2|




When x<1, we have f(x) = -2x+3, which is a polynomial function and polynomial function is continuous everywhere.


When 1≤x<2, we have f(x) = 1, which is a constant function and constant function is continuous everywhere.


When x≥2, we have f(x) = 2x-3, which is a polynomial function and polynomial function is continuous everywhere.


Hence, f(x) = |x-1| + |x-2| is continuous everywhere.


Let’s discuss the differentiability of f(x) at x=1 and x=2.


We have




Lf’(1) =


=


= ( f(x) = -2x+3, if x< 1)


=


=


Rf’(1) =


= ( f(x) = 1, if 1≤ x< 2)


=0


Lf’(1) ≠ Rf’(1)


f(x) is not differentiable at x=1.


Lf’(2) =


= (f(x) = 1, if 1≤ x< 2 and f(2) = 2×2-3 =1 )


=0


Rf’(2) =


=


= ( f(x) = 2x-3, if x≥ 2)


=


=


Lf’(2) ≠ Rf’(2)


f(x) is not differentiable at x=2.


Thus, f(x) = |x-1| + |X-2| is continuous everywhere but fails to be differentiable exactly at two points x=1 and x=2.


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