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How do you find the domain and range of #f(x) = 1/(1+x^2)#?

Answer 1

Isaiah Anthony

The domain is #x in RR#. The range is #y in (0,1]#

The denominator is #=1+x^2#

#AA x in RR#, #1+x^2>0#

Consequently,

The domain of #f(x)# is #x in RR#

To find the range, follow these steps:

#y=1/(1+x^2)#

#y(1+x^2)=1#

#y+yx^2=1#

#yx^2=1-y#

#x^2=(1-y)/y#

#x=sqrt((1-y)/y)#

The range of #f(x)# is the domain of #x#

#((1-y)/y)>0#

#y in RR _+^("*")#

#1-y>=0#

#y<=1#

Consequently,

The range is #y in (0,1]#

graph{1/(1+x^2) [-5.625, 5.625, -11.25, 11.25]}

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Answer 2

Eli Assouline

Domain: All real numbers.
Range: ( (0, 1] ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.