Q. 975.0( 1 Vote )

# Fill in the

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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