anonymous
  • anonymous
Find the domain and range of: f(x)=18sqrt(x)-17
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
D={x: where x belong to R and x >=17} R={ y>=0}
anonymous
  • anonymous
so, would it be 18x-17>=0? and solve?
anonymous
  • anonymous
is this your function: \(\large f(x)=18\sqrt{x}-17 \)

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anonymous
  • anonymous
yes
anonymous
  • anonymous
|dw:1346012919057:dw|
anonymous
  • anonymous
the domain of f is all the x values that give f(x) a real value.... so looking at the function, the function will not give you a real number when x is less than 0....
anonymous
  • anonymous
so the actual question in finding the domain is "what x values is \(\large \sqrt x =real\) number
anonymous
  • anonymous
would I do.. 18x-17<0 and solve?
anonymous
  • anonymous
x>=-17
anonymous
  • anonymous
root of x is always larger than 0
anonymous
  • anonymous
You're just looking to be sure that what's under the root does not go negative. So x cannot be negative. Since the smallest value of x is 0 (because of the domain), then the range's minimum value will happen when x is 0. (So plug in 0 for x, and see what y is). The upper limit of the function will go on forever.
anonymous
  • anonymous
-17?
anonymous
  • anonymous
You got it, that's the minimum of the range. As x gets bigger and bigger, there is no limit, it just grows and grows forever.
anonymous
  • anonymous
so.. [-17,infinity) ?
anonymous
  • anonymous
That is your range, yes.
anonymous
  • anonymous
how do I find the domain with this function?
anonymous
  • anonymous
(-inf,inf) ?
anonymous
  • anonymous
Well, as we talked about earlier, the domain would be anything that is a "no no". You have a square root, and here, you cannot take the square root of a negative number, as it has no real solutions.
anonymous
  • anonymous
anything that is NOT a "no-no", sorry ;]
anonymous
  • anonymous
1?
anonymous
  • anonymous
No, you simply have sqrt(x), since you cannot take the square root of a negative number, we know that x MUST be greater than or equal to 0.
anonymous
  • anonymous
so.. 17?
anonymous
  • anonymous
|dw:1346014535907:dw|
anonymous
  • anonymous
We're looking for x's that might make the equation undefined. That's what it means to find the domain. The only thing in this equation that can become undefined (or have no real solutions) is the square root. Which can't be negative.
anonymous
  • anonymous
so x is all real numbers that are positive then?
anonymous
  • anonymous
Yep, or in other words x >= 0
anonymous
  • anonymous
how do I put that into interval notation though? [0,inf)

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