# Find set of numbers where f(x)= arctg(x/sqrt(3)) for |x|≤1, or (π/6)*sgn(x)+1/2*(|x|-1) for |x|>1 , is continuous.

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### 1 Answer

You need to notice that the function has the expression `f(x) = arctan(x/sqrt 3)` if -`1<=-x<=1` and it has the expression `f(x) = (pi/6)*sgn(x)+1/2*(|x|-1)` if `x<-1` or `x>1` such that:

`f(x) = {(arctan(x/sqrt 3), -1<=-x<=1), ((pi/6)*sgn(x)+1/2*(|x|-1), x<-1 or x>1):}`

You need to check if the given function is continuous at the points `x = 1` and `x = -1` such that:

`f(1) = arctan(1/sqrt3) => f(1) = pi/6`

`f(-1) = arctan(-1/sqrt3)`

You need to use the following trigonometric identity, such that:

`arctan(-x) = -arctan x`

Reasoning by analogy yields:

`arctan(-1/sqrt3) = -arctan(1/sqrt 3) = -pi/6`

You need to evalaute the side limits to the point `x = -1` , such that:

`lim_(x->-1, x<-1)(pi/6)*sgn(x)+1/2*(|x|-1) = -pi/6 + (1/2)(|-1| -1)`

`lim_(x->-1, x<-1)(pi/6)*sgn(x)+1/2*(|x|-1) = -pi/6 + (1/2)(1 - 1)`

`lim_(x->-1, x<-1)(pi/6)*sgn(x)+1/2*(|x|-1) = -pi/6`

`lim_(x->-1, x>-1)(arctan(x/sqrt3)) = arctan(-1/sqrt 3) = -pi/6`

Notice that the values of side limits are equal and they are also equal to the value of the function at the point `x = -1` , hence, the function f(x) is continuous at `x = -1` .

You need to check if the function is also continuous at `x = 1,` hence, you need to evaluate the side limits of the function at`x = 1,` such that:

`lim_(x->1, x<1) (arctan(x/sqrt3)) = arctan(1/sqrt3) = pi/6`

`lim_(x->1, x>1)(pi/6)*sgn(x)+1/2*(|x|-1) = (pi/6)*sgn(1)+1/2*(|1|-1) `

`lim_(x->1, x>1)(pi/6)*sgn(x)+1/2*(|x|-1) = (pi/6) + 1/2*(1-1)`

`lim_(x->1, x>1)(pi/6)*sgn(x)+1/2*(|x|-1) = pi/6`

Notice that the values of side limits are equal and they are also equal to the value of the function at the point `x = 1` , hence, the function `f(x)` is also continuous at `x = 1` .

**Hence, evaluating the continuity of the given function at the points `x = -1` and `x = 1` , yields that the function is continuous at points `x = -1` and `x = 1` .**