Find the domain and range of:
f(x)=18sqrt(x)-17

- anonymous

Find the domain and range of:
f(x)=18sqrt(x)-17

- katieb

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- anonymous

D={x: where x belong to R and x >=17}
R={ y>=0}

- anonymous

so, would it be 18x-17>=0? and solve?

- anonymous

is this your function: \(\large f(x)=18\sqrt{x}-17 \)

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## More answers

- anonymous

yes

- anonymous

|dw:1346012919057:dw|

- 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

so the actual question in finding the domain is "what x values is \(\large \sqrt x =real\) number

- anonymous

would I do.. 18x-17<0 and solve?

- anonymous

x>=-17

- anonymous

root of x is always larger than 0

- 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

-17?

- 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

so.. [-17,infinity) ?

- anonymous

That is your range, yes.

- anonymous

how do I find the domain with this function?

- anonymous

(-inf,inf) ?

- 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

anything that is NOT a "no-no", sorry ;]

- anonymous

1?

- 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

so.. 17?

- anonymous

|dw:1346014535907:dw|

- 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

so x is all real numbers that are positive then?

- anonymous

Yep, or in other words x >= 0

- anonymous

how do I put that into interval notation though? [0,inf)

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