what is the domain of f(x) = abs((9-x^2)/(x+3))?

Question

my teacher told me to find the domain symbolically and find the range using the calculator.

Answer 1

You can take the absolute value of any number; however, you cannot divide by 0.
Set the denominator equal to zero and solve for x.
x +3 = 0
x = -3
Domain = all real numbers except -3.
For the range,
The absolute value is never negative, so you can automatically restrict the range to y >= 0 (all numbers greater than or equal to zero).
Now are there any other numbers that y cannot be?
No, you can check this using your calculator.

Answer 2

Well Aditya . i would tell you how you can find Domain and range.
well dear first you need to simplify this function .

Step 1;

f(x) = abs((9-x^2)/(x+3))

first simplify this one " (9-x^2) "
(9-x^2) = ( (-x^2) + 9) = ( -x -3)(x - 3)

explanation:
{ ( -x -3)(x - 3) = (-x * +x) + (-x *-3) +(-3*x )(-3*-3)=-x^2 +3x - 3x + 9 = -x^2 + 9 , so now you know how we could simplify this function}.

now lets get back;

f(x) = | (( -x -3)(x - 3)) / (x +3) | = now we can factor from " -1 " ,so we have; f(x) =| -(( x +3)(-x + 3)) / (x +3) | ; now we can remover " (x+ 3) ",then
f(x) would be ; f(x) =| -(-x + 3) | ; f(x) =| (x - 3) |
Now Domain can be anything ,i mean any number positive or negative and zero

and range ;

if x= -1 ;y or f(x) = | -1 -3| = |-4| = +4
if x=0 ;y or f(x) = | 0 -3| = |-3| = +3
if x=+1 ;y or f(x) = | +1 -3| = |-2| = +2

so you the result is just POSITIVE NUMBERS
so it means
Range=all positive numbers .

Answer 3

Domain is possible x values

Range is possible y values

Just think about it alphabetically, x comes before y and D(omain) comes before R(ange).

Now absolute value makes the equation positive no matter what, so the Range is all positive numbers (not sure if it includes 0 too, just ask, I think it does include 0) and Domain is all real numbers except 3 and -3 because 3 causes the equation to be zero divided by a number, which is impossible.

Answer 4

x cannot be -3. The procedure 'cancel out' is not maths. It is what seems to be but not the real procedure. The real procedure is multiplying both numerator and denominator by the same number except zero, thus the fraction remains the same. E.g. 3/5 = 6/10 = 600/1000 but 3/5 is not equal to (3x0)/(5x0) = 0/0 which is undefined!

So, f(x) = abs(3-x) is only true for x not equal to -3.

Answer 5

f(x) = |(9 - x^2)/(x + 3)|

f(x) = |((3 - x)(3 + x))/(x + 3)|
f(x) = |3 - x|

x <= 3
x >= 3

Answer 6

all but -3 is the answer...

Answer 7

who could answer is really smart cuz i have problem with math