Domain and range question, HELP?

Question

State the domain and range of f(x) = x/√1 – x^2.


Note: 1 – x^2 is under the square root sign

Answer 1

A function is defined from a set, the domain, to a set, the range. It is not necessarily true that every point in the range is the value of the function, so it seems more likely that you are asking about the image of the function, which is the set of values it takes.

Okay, with that out of the way, we can begin to look at values of x for which this makes sense. What is needed is that 1-x^2 > 0, since square roots of negative numbers are invalid in this context, and 0 will produce division by zero. That is all x with -1 < x < 1.

The image, all values that the functioin takes on, is the set of all real numbers, and this can be see easily since for x near 1, the value of the function becomes large, and for x near -1 the value becomes large negative.

Answer 2

to find the domain put the denominator >0 bcoz the denominator conatin radical function ,if it contain normal function put it =0

so
y = x / √(1-x²)
(1-x²) >0
(1-x)(1+x)>0
x<1 ,x>-1
so the domain between ]1,-1[

to find the range u have to convert the function from y= x.....
to x= y....
so,we have y= x/√(1 – x^2)

y² = x² /(1-x²)

y²(1-x²) =x²
y² -y²x² =x²
y²= x² + y²x²
y² =x² (1+y²)
y² /(1+y²) =x²

x² = y² /(1+y²)

x = y / √(1+y²) now take the domain for the new function and thats domain will be the range for the original function
so
the domain is :
(1+y²) > 0
y² > -1 ,so it's impossible to take the sqrt for both sides since there is no sqrt(-1)
so,the domian for new function is all real numbers.
And
the range for the original function is all real numbers too.

i hope this help

Answer 3

I suppose you only want real values. Snce the sqrt is in the denominator , it can't be 0 and f is defined for every x such that 1 - x^2 >0, so x in (-1, 1). This the domain.

For the range, observe that f is continuous on it's domain and f(0) = 0. As x -> 1, x -> 1 and sqrt(1-x^2) -> 0, so that f(x) -> oo. Since f is continuous, on [0,1) it takes on every value on [0, oo). And since f is an odd function, (f(-x) = -f(x) for every x in (-1,1)) om (-oo, 0] it takes all values on (-oo, 0]. Thefore, the range of f is the whole real line.

Answer 4

domain
1-x^2>0
x^2<1
-1 range
-infinity to + inf