The given function is

The function f is defined at all points of the interval [0,10].

Let k be the point in the interval [0,10].

Then, we have 5 cases i.e., 0≤ k < 1, k = 1, 1 < k < 3, k = 3 or 3 < k ≤ 10.

Now, Case I: 0≤ k < 1

Then, f(k) = 3

= 3= f(k)

Thus,

Hence, f is continuous in the interval [0,10).

Case II: k = 1

f(1) = 3

= 3

= 4

Hence, f is not continuous at x = 1.

Case III: 1 < k < 3

Then, f(k) = 4

= 4 = f(k)

Thus,

Hence, f is continuous in (1, 3).

Case IV: k = 3

= 4

= 5

Hence, f is not continuous at x = 3.

Case V: 3 < k ≤ 10

Then, f(k) = 5

= 5 = f(k)

Thus,

Hence, f is continuous at all points of the interval (3, 10].

Therefore, x = 1 and 3 are the points of discontinuity of f.