How do you determine if #f(x) = ln(x + sqrt(x^2 + 1))# is an even or odd function?
1 Answer
Jul 30, 2016
Explanation:
-
An even function is one for which
#f(-x) = f(x)# for all#x# in the domain. -
An odd function is one for which
#f(-x) = -f(x)# for all#x# in the domain.
Note that:
#(sqrt(x^2+1)+x)(sqrt(x^2+1)-x) = (x^2+1)-x^2 = 1#
So we find:
#f(-x) = ln(-x + sqrt(x^2+1))#
#= ln(1/(x+sqrt(x^2+1)))#
#= -ln(x + sqrt(x^2+1)) = -f(x)#
So