# Given function f(x)={x}(1-{x}),determine interval of continuity if domain is [0,3]?

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You need to use the formula of fractional part, such that:

`{x} = x - [x]`

You should discuss the function over the following intervals, such that:

`x in [0,1) => {x} = x - 0 => {x} = x `

Replacing `x` for `{x}` in equation of function yields:

`f(x) = x(1 - x)`

`x in [1,2) => {x} = x - 1`

Replacing `x - 1` for `{x}` in equation of function yields:

`f(x) = (x - 1)(1 - x + 1) => f(x) = (x - 1)(2 - x)`

`x in [2,3) => {x} = x - 2`

Replacing `x - 2` for `{x}` in equation of function yields:

`f(x) = (x - 2)(1 - x + 2) => f(x) = (x - 2)(3 - x)`

Hence, you need to test if the function f(x) = `f(x) = {(x(1 - x)),((x - 1)(2 - x),),((x - 2)(3 - x)):}` is continuous over the interval `[0,3]` .

You need to test the continuity at `x = 1` such that:

`lim_(x->1,x<1) x(1 - x) = lim_(x->1,x>1)(x - 1)(2 - x) => 1(1 - 1) = (1 - 1)(2 - 1) => 0 = 0`

Hence, the function is continuous at` x = 1` .

You need to test the continuity at `x = 2` such that:

`lim_(x->2,x<2) (x - 1)(2 - x) = lim_(x->2,x>2) (x - 2)(3 - x)`

`(2 - 1)(2 - 2) = (2 - 2)(3 - 2) => 0 = 0`

Hence, the function is continuous at `x = 2` .

**Hence, testing if the function is continuous over its domain `[0,3]` yields that the statement is valid.**