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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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 entire section contains 2 answers and 226 words.)
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calendarEducator since 2010
write12,551 answers
starTop subjects are Math, Science, and Business
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}.
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