HOW CAN I FIND THE DOMAIN AND THE RANGE of these functions?

Question

Please help me. If I see some examples on how to solve these, maybe I'll get better at them.

12x^2–15

DOMAIN: (-infinity, infinity)
RANGE: ???????

sqrt(x) + 12

DOMAIN: ??????
RANGE: ??????

x+12/x^2-36

That's supposed to be a fraction.

DOMAIN: ?????????

I have to write all of these in interval notation. These are from work online, which is completely different from my notes and the problems from my text book, leaving me very, very confused. Please help me.

Answer 1

In the first question you have to find the minimum value of 12x^2-15, in the second the domain is clearly where the square root is defined, i.e. x>=0; for the range you to look for the minimum of sqrt(x)+12, which is very feasible :)

In the third question all depends if you have some more brackets in the denominator, as in
(x+12)/(x^2-36)
or not. Anyway you have to take away all points x which give a denominator =0

Answer 2

The 3 problems you're asking about are typically called functions of an independent variable (in your cases x is the symbol chosen to represent the independent variable).

The domain of the independent variable is all real values that "kick out" a function value.

So in f(x) = 12x^2 - 15 you look at the operation(s) going on
(only multiplying and subtraction, which are "closed" -- that is, any mult and subt operation with real numbers will always have real number answers), and thus conclude ALL real numbers, i.e. (-inf, +inf) is the domain.

Having determined the domain then you determine what function values (range elements) are gonna be generated by elements from the domain. You notice that squaring any real number will make it always >= 0, then multiplied by 12 and reduced by 15. Logically, then the range has a lower limit of -15, and then can get as large as you please. So the range is [-15, +inf)

Second problem: f(x) = sqrt(x) + 12

You only take sqrt of non-neg numbers and your sqrt is single valued, always >= 0. Your domain is [0, + inf).
Your range is [12, +inf)
-------------------------------------------------
A "raw spot" in your problem f(x) = x + 12/x^2 - 36 is that without grouping symbols there tends to be ambiguity. If you feed f(0) = 0 + 12/0^2 - 36 into a scientific calculator, you'll get a divide by zero error (implying zero isn't in the domain). But IF YOU MEANT

f(x) = x + 12/(x^2 -36), then f(0) would be -1/3 and zero would be in the domain.

The point is that with the ambiguity in your 3rd problem it's hard to determine what the function is and therefore what its domain and range would be. PLEASE USE GROUPING SYMBOLS TO REMOVE ANY AMBIGUITY.

Answer 3

Remember - the domain is all possible x values, the range is all possible y values.

----------------------------
12x^2 - 15

the domain is all real numbers (- infinity, infinity)
The range is ( -15, infinity)
------------------------
sqrt (x) + 12

the domain is all positive real numbers (0, infinity) - because you can't have the square root of a negative number in this situation.

The range is all positive real numbers (12, infinity) - it's "positive" square root x, so you'll always have a positive number plus 12, so it's always going to be greater than 0.
--------------------------------------------------
(x+12)/(x^2 - 36)

First, set the denominator =0 and solve for x:
x^2 - 36 = 0
(x - 6)(x + 6) = 0
x = -6 and 6

So, the domain is all real numbers
except x is NOT equal to -6 or 6.
The range will be all real numbers (-infinity, infinity)

Answer 4

12x^2-15
range: (-15, infinity)

sqrt(x) +12
domain: x>or= 0 or (0, infinity)
range: (12, Infinity)

(x+12)/ (x^2 -36)
domain: x not = 6 and x not= -6 or (-infinity,-6) and
(-6,6) and (6, infinity)

Answer 5

12x^2-15
Domain (-infinity , +infinity)
range [-15 , +infinity)

sqrt(x)+12
Domain [0 , +infinity)
Range [12 , +infinity)

(x+12)/x^2-36
Domain (-infinity , 6) + (6 , +infinity) (Any number exclude 6)