Ask your own question, for FREE!
Calculus1 13 Online
OpenStudy (anonymous):

find domain and range of f(x)=ln(x+sqrt x^2 + 1 )

OpenStudy (campbell_st):

well with log function the basic domain starting point is log(x) x> 0 so you need to look at the values of x that make \[x + \sqrt{x^2 + 1} > 0\]

OpenStudy (anonymous):

yea and need answers for this Q and do u know answer?

OpenStudy (campbell_st):

well it's reasonably straight forward \[\sqrt{x^2 + 1} > -x\] squaring any value will make it positive... then adding 1 will always make it larger than -x so what domain do you think this covers

OpenStudy (campbell_st):

one method you might consider is graphing the function https://www.desmos.com/calculator

OpenStudy (anonymous):

okay i get answers from website you gave me and thnx

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!