Algebraically determine the domain and range:

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Algebraically determine the domain and range:

Mathematics
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Typing the equation now
Sure, go ahead;0
|dw:1440467179872:dw|

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Other answers:

Equation button isn't working
You can copy paste my code (it works in text) `\(\large\color{black}{ \displaystyle y=\sqrt{x^2-4x-5} }\)`
\(\large\color{black}{ \displaystyle y=\sqrt{x^2-4x-5} }\)
So, you know that the part inside the square root must not be negative, right?
Yes
There we go, had to restart my browser
So, set: \(\large\color{black}{ \displaystyle x^2-4x-5\ge0 }\) and solve for x
this will give you the domain
What about the range? That's what really always confuses me
Well, \(\large\color{black}{ \displaystyle x^2-4x-5 }\) will equal to 0 for at least 1 value of x, and if \(\large\color{black}{ \displaystyle x^2-4x-5=0}\) for some x-value, then we know that y is =0 for some x-value. Now, you can probably tell answer these two questions without me: ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` 1) Will you ever get a negative output for y? -------------------------------------------------- 2) \(\large\color{black}{ \displaystyle x^2-4x-5 }\) can get as large as you want, \(\\[0.8 em]\) and thus \(\large\color{black}{ \displaystyle \sqrt{x^2-4x-5} }\) can also get as large as you want. \(\\[0.9 em]\) Knowing this, is there a limit for how large the range (the output) can be?
1) no 2) no
Yes, correct...
So can you tell me the range ?
(if you got questions, it's ok, ask)
|dw:1440467760532:dw|
Yes, that is right
Is that right?
yes
Ah ok
Thanks
Yes, you are welcome!
And the domain is?
|dw:1440467835822:dw|
\(\large\color{black}{ \displaystyle y=\sqrt{x^2-4x-5} }\) You know that the part inside the square root can't be negative: So we set: \(\large\color{black}{ \displaystyle x^2-4x-5\ge0 }\) \(\large\color{black}{ \displaystyle (x-5)(x+1)\ge0 }\) For left side to be greater than 0, factors have to be either both negative or both positive or one of factors has to be equal to 0. \(\large\color{black}{ \displaystyle x\le-1 }\) then for values less than -1, both factors are negative (and - times - =+ , and +>0) Or when x=-1 one factor is zero, and 0\(\ge\)0 which also satisfies the equation. \(\large\color{black}{ \displaystyle x\ge5 }\) when x=5 we again get \(0\le0\) (which satisfies the equation). For values greater than 5, both factors are positive (and + times + = +, which is greater than 0 as well)
Whoops, got the signs mixed
However if you said \(x\ge-1\) (you made that little mistake) then any number between -1 and 5 will disprove the statement. For example x=0, \(\large\color{black}{ \displaystyle (x-5)(x+1)\ge0 }\) \(\large\color{black}{ \displaystyle (0-5)(0+1)\ge0 }\) \(\large\color{black}{ \displaystyle (-5)(1)\ge0 }\) \(\large\color{black}{ \displaystyle -5\ge0 }\) and that is not true.
So, as I showed, the solutions are \(x\le -1\) \(x\ge 5\)
Thanks again!
Yw

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