What is the domain and range of #ln(x^2+1)#?

Answer 1

Domain is #RR#+, Range is #RR^+#

Domain is given by #x^2 +1# >0. That means all real values of x, that is, it would be #RR#
For range, exchange x and y in #y= ln (x^2+1)# and find the domain. Accordingly, #x= ln(y^2 +1)#
#y^2= e^x-1#. The domain of this function is all #x>= 0# that means all real numbers #>=#=0
Hence the range of given function would be all Real numbers #>=0#
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Answer 2

The domain of ln(x^2 + 1) is all real numbers such that x^2 + 1 is greater than zero, or x^2 + 1 > 0. Solving this inequality yields the domain as (-∞, ∞). The range of ln(x^2 + 1) is all real numbers, or (-∞, ∞).

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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.

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.

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.

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.

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