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

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Answer 1 · 2016-05-31T18:04:40

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 $>=0$

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