1. ## Help please a question

Hey. Can someone help me with these questoins

1. What is the domain of h(x)= square root of x squared - 4x - 5

How do you find the domain of things I know it's X but how do you find it

2. Originally Posted by key
1. What is the domain of h(x)= square root of x squared - 4x - 5
We require that $\displaystyle x^2-4x-5\ge0\implies(x+1)(x-5)\ge0.$

Do you know how to tackle that?

3. Originally Posted by key
Hey. Can someone help me with these questoins

1. What is the domain of h(x)= square root of x squared - 4x - 5

How do you find the domain of things I know it's X but how do you find it
Well, you can't take the square root of a negative number , so everything inside the square root, must be positive. (if you have difficulty seeing this, ask yourself what real number multiplied by itself will return a negative number? for example what number squared would be negative 1? There is not one, because a negative number squared will become positive, and a positive number squared will remain positive)

So
$\displaystyle 0 \leq x^{2}+2x5$

Now, if h(x) is positive, but was negative (or vice versa), it must have been equal to zero at some point, because zero is where signs change. So lets find where this equation equals zero.
$\displaystyle 0 = x^{2}-4x-5$

Factor into (x-1) and (x+1)
$\displaystyle 0 = (x-5)(x+1)$

so $\displaystyle x-5=0 ~~~\Longrightarrow ~~x=5$
and $\displaystyle x+1=0 ~~\Longrightarrow ~~x=-1$

Now, we know where the function equals zero, which is where it changes signs. We just need to check the intervals.

The intervals are $\displaystyle (-\infty,-1), (-1,5), \mbox{ and } (5,\infty)$

So lets pick the point -2 to check the sign of the first interval, 0 to check the sign of the second interval, and 6 to check the sign of the third interval.

let $\displaystyle g(x) = x^{2}-4x-5$
$\displaystyle g(-2)=7 ~~\Longrightarrow ~~(-\infty,-1]$ is in the domain
$\displaystyle g(0)=-5 ~~\Longrightarrow ~~(-1, 5)]$ is not in the domain
$\displaystyle g(6)=~7 ~~~\Longrightarrow ~~[5, \infty)$ is in the domain

(note, we used brackets on -1 and on 5, because they are roots, which means they cause the equation to equal zero, and we can take the square root of zero, so they are in the domain)

So the domain is: $\displaystyle (-\infty, -1]\cup[5, \infty)$

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Now you try for $\displaystyle h(x)=\sqrt{-x^{2}-2x+8}$