Find the domain of the function f(x) = √(x²- 4x -5)

Solution:  Since this function contains a square root, the quantity under the radical must be positive or possibly zero for defined functional values. So, we need to find x-values for which this holds. Let’s take it in sections.

For x²- 4x -5 = 0, we need the factors of the left side. So, we solve (x-5)(x+1) = 0.

Setting each factor equal to 0, we get:  x-5=0, x+1=0 → x=5, x= -1. These values will be part of the domain.

For x²- 4x -5 to be positive, we need to solve the quadratic inequality x²- 4x -5 >0

or, in factored form,   (x-5)(x+1) > 0. This will be satisfied when both factors are positive or both factors are negative. Let’s take them separately.

Both factors are positive: x-5>0 and x+1>0.  Solving for x in each, we get, x>5 and x>-1. The only way both can be true is when x>5. These values for x will be part of the domain as well. (i.e., if a number is >5 it is automatically >-1, but if a number is x>-1 it is NOT automatically >5, example, 2).

Both factors are negative: x-5<0 and x+1<0.  Solving for x in each, we get, x<5 and x<-1. The only way both can be true is when x<-1. These values for x will be part of the domain as well. (i.e., if a number is <-1 it is automatically <5, but if a number is <5 it is NOT automatically <-1, example, 4).

Bringing all parts together, we have   x ≤ -1   OR   x ≥ 5 for the full domain. We can also express these values using interval notation and set union by   (-∞;-1] U [5;∞).

Note:  Be careful,   stating   -1≤ x ≥ 5 makes no sense since inequalities can only be strung together with an “and” situation (i.e.,   x>2 and x<5 may be written as   2<x<5).