# Find the domain and range of the following: y = x^2 , y = sqrt(1 – x^2), y = 1/x, y = sqrt(x) , y = sqrt(4-x).

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For a function y = f(x), all the values that x can take form the domain and all the values y can take form the range.

Here, I have assumed that y and x can only take on real values.

For y = x^2, the domain is –inf < x < +inf and the range is y>=0.

For y = sqrt (1 – x^2), the domain is 1<= x <= 1 and the range is 0 <= y <=1

For y = 1/x, the domain is all values except x = 0, and the range is all values except y =0.

For y = sqrt(x), the domain is 0<= x, and the range is 0 <= y

For y = sqrt (4-x), the domain is x <=.4 and the range is 0 <= y.

y= x^2

The domain is all real numbers such that:

x = (-inf, inf)

Now to notice that y is a positive number . Then the ramge is:

y= [ 0, +inf)

y = sqrt(1-x^2)

The domain is all x values such that 1- x^2 >= 0

==> -x^2 > = -1

==> x^2 = < 1

==> x =< 1 and x >= -1

==>the somain is [ -1, 1]

The range is : ( 0, 1]

y= sqrtx

x >= 0 ==> the domain is [ 0, inf)

==>The range is : y= [ 0, inf)

y= sqrt(4-x)

The domain is 4-x > = 0

==> 4 > = x

==> x =< 4

Then the domain is ( -inf, 4]

Then the range will be [ 0, inf)

1)y = x^2.

The domain is the set of all x values for which the values of y is rea. Here x can take all real values. So the domain of y = x^2 is( -infinity < x < infinity). The range of the function is the set of values y takes: y > =0. Or 0 = < y < infinity.

2) y = sqrt(1-x^2).

Domain of the function: Since sqrt(1-x)^2 is real only if 1-x^2 > 0. Or x^2 < 1, which is possible only if -1= < x x < = 1. Therefore the domain of the function is the set of values of x in the interval (-1 , 1). Therefore y = x^2 >= 0 and y < = 1. So the range of the funtion is the set of y values in the interval (0 ,1).

3) y = 1/x,

When x = 0, y is indeterminate . So x can teke all values except 0. So the domain is the set {-infinity , infinity}- {0}.

y can take all vallues but not zero . Therefore the range is {-infinity , infinity }- {0}.

4)y = sqrt(x) .

x cannot take negative values as y becomes not real when x < 0. So , the domain of the function is x >=0.

Range of the function is the set {y: y> 0 }.

5) y = sqrt(4-x).

4-x> 0 as y will not bereal for 4-x< 0.

Therefore the domain is the set of of x x values for which x > =4.

Domain of the function = {x : x> =0}.

Range is the set of all possible y values or values of sqrt(4-x).

So range of the function = {y : y > 0}.