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monroe17
Find the domain and range of: f(x)=18sqrt(x)-17
D={x: where x belong to R and x >=17} R={ y>=0}
so, would it be 18x-17>=0? and solve?
is this your function: \(\large f(x)=18\sqrt{x}-17 \)
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....
so the actual question in finding the domain is "what x values is \(\large \sqrt x =real\) number
would I do.. 18x-17<0 and solve?
root of x is always larger than 0
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.
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.
That is your range, yes.
how do I find the domain with this function?
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.
anything that is NOT a "no-no", sorry ;]
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.
|dw:1346014535907:dw|
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.
so x is all real numbers that are positive then?
Yep, or in other words x >= 0
how do I put that into interval notation though? [0,inf)