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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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hala718 eNotes educator | Certified Educator

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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)

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neela | Student

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}.