Question

Find the domain and range of:

f(x)=18sqrt(x)-17

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

yes

Answer 5

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....

Answer 6

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

Answer 7

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

Answer 8

x>=-17

Answer 9

root of x is always larger than 0

Answer 10

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.

Answer 11

-17?

Answer 12

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.

Answer 13

so.. [-17,infinity) ?

Answer 14

That is your range, yes.

Answer 15

how do I find the domain with this function?

Answer 16

(-inf,inf) ?

Answer 17

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.

Answer 18

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

Answer 19

1?

Answer 20

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.

Answer 21

so.. 17?