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JacksonJRB
 one year ago
Algebraically determine the domain and range:
JacksonJRB
 one year ago
Algebraically determine the domain and range:

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JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0Typing the equation now

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Sure, go ahead;0

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440467179872:dw

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0Equation button isn't working

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1You can copy paste my code (it works in text) `\(\large\color{black}{ \displaystyle y=\sqrt{x^24x5} }\)`

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle y=\sqrt{x^24x5} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1So, you know that the part inside the square root must not be negative, right?

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0There we go, had to restart my browser

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1So, set: \(\large\color{black}{ \displaystyle x^24x5\ge0 }\) and solve for x

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1this will give you the domain

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0What about the range? That's what really always confuses me

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Well, \(\large\color{black}{ \displaystyle x^24x5 }\) will equal to 0 for at least 1 value of x, and if \(\large\color{black}{ \displaystyle x^24x5=0}\) for some xvalue, then we know that y is =0 for some xvalue. Now, you can probably tell answer these two questions without me: ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` 1) Will you ever get a negative output for y?  2) \(\large\color{black}{ \displaystyle x^24x5 }\) can get as large as you want, \(\\[0.8 em]\) and thus \(\large\color{black}{ \displaystyle \sqrt{x^24x5} }\) 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?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1So can you tell me the range ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1(if you got questions, it's ok, ask)

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440467760532:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yes, that is right

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yes, you are welcome!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1And the domain is?

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440467835822:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle y=\sqrt{x^24x5} }\) You know that the part inside the square root can't be negative: So we set: \(\large\color{black}{ \displaystyle x^24x5\ge0 }\) \(\large\color{black}{ \displaystyle (x5)(x+1)\ge0 }\) 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. \(\large\color{black}{ \displaystyle x\le1 }\) 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)

JacksonJRB
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, got the signs mixed

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1However if you said \(x\ge1\) (you made that little mistake) then any number between 1 and 5 will disprove the statement. For example x=0, \(\large\color{black}{ \displaystyle (x5)(x+1)\ge0 }\) \(\large\color{black}{ \displaystyle (05)(0+1)\ge0 }\) \(\large\color{black}{ \displaystyle (5)(1)\ge0 }\) \(\large\color{black}{ \displaystyle 5\ge0 }\) and that is not true.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1So, as I showed, the solutions are \(x\le 1\) \(x\ge 5\)
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