# What is domain of arctg `sqrt(x^2-5x+4)`

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The domain of the arctangent function contains all real numbers, hence, you need to consider the following condition for the principal argument of the function, such that:

`sqrt(x^2 - 5x + 4) in R`

Since the square root holds only if the radicand is positive, you need to consider the following condition, such that:

`x^2 - 5x + 4 >= 0`

You need to attach the quadratic equation, such that:

`x^2 - 5x + 4 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (5+-sqrt(25 - 16))/2 => x_(1,2) = (5+-3)/2`

`x_1 = 4; x_2 = 1`

The quadratic is positive or zero if `x in (-oo,1]U[4,oo).`

**Hence, evaluating the domain of arctangent function yields that it coincides with the domain of radical function, thus **`x in (-oo,1]U[4,oo).`

**Sources:**