What's the domain of $f(x) = \sqrt{x^2 - 4x - 5}$? What's the domain of the function $f(x) = \sqrt{x^2 - 4x - 5}$ ?
Thanks in advance.
 A: The function $x\mapsto\sqrt x$ is defined on the interval $[0,+\infty)$ so the given function $f$ is defined for $x$ such that $x^2-4x-5\ge0$.
The reduced discriminant $\Delta'=9$ hence the roots are: $x_1=-1$ and $x_2=5$ and then $f$ is defined on $(-\infty,-1]\cup[5,+\infty)$.
A: $f(x) = \sqrt{x^2 - 4x - 5}$
Since the square root of a negative number is imaginary, the condition is that :
$$x^2 - 4x - 5 \geq 0$$
$$ (x-2)^2 - 9 \geq 0$$
$$ (x-2) \geq 3$$ or
$$ (x-2) \leq -3$$
Therefore the domain of the function $f(x)$ would be:
$$(-\infty, -1] \cup [5, +\infty)$$
Hope the answer is clear ! And wish you a happy new year !
A: The domain is $\text{domain}(f)\geq 0$, because since you are dealing with the reals, it cannot be negative inside the square root. 
To determine domain, solve: $x^2-4x-5=0$: 
$(x-5)(x+1)=0 \implies x=5,-1$ These are the roots, so the domain is: $(-\infty,-1] \cup [5,\infty)$.  
A: The domain is $D =\{ x \in \mathbb{R} | x^{2}-4x-5 \geq 0 \}$.
The roots of $x^{2}-4x-5$ are $-1$ and $5$ so $D =(-\infty, -1] \cup [5, +\infty)$.
A: For the function $f$ you gave to be undefined, the input to the square root must be negative. First, notice that 
$$
x^2-4x-5=(x+1)(x-5)
$$
So for this to be negative one of the terms must be negative and the other must be positive. So what numbers satisfy $x+1>0$ and $x-5<0$ or $x+1<0$ and $x-5>0$?
