Q. 73.5( 66 Votes )

# Find all points of discontinuity of f, where f is defined by

Answer :

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 5 cases i.e., k < -3, k = -3, -3 < k < 3, k = 3 or k > 3

Now, __Case I:__ k < -3

Then, f(k) = -k + 3

= -k + 3= f(k)

Thus,

Hence, f is continuous at all real number x < -3.

__Case II:__ k = -3

f(-3) = -(-3) + 3 = 6

=-(-3) + 3 = 6

= -2×(-3) = 6

Hence, f is continuous at x = -3.

__Case III:__ -3 < k < 3

Then, f(k) = -2k

= -2k = f(k)

Thus,

Hence, f is continuous in (-3,3).

__Case IV:__ k = 3

= -2×(3) = -6

= 6 × 3 + 2 = 20

Hence, f is not continuous at x = 3.

__Case V:__ k > 3

Then, f(k) = 6k + 2

= 6k + 2= f(k)

Thus,

Hence, f is continuous at all real number x < 3.

Therefore, x = 3 is the only point of discontinuity of f.

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