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

Question

campbell_st · Answer

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  

$$x + \sqrt{x^2 + 1} > 0$$

anonymous · Answer

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

campbell_st · Answer

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

campbell_st · Answer

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

anonymous · Answer

okay i get answers from website you gave me and thnx