What is the domain and range of # y=sqrt(x^2-1)#?

Question

Eva Adamson · Accepted Answer

Domain: #(-oo, -1] uu [1, + oo)#
Range: #[0, + oo)# The requirement that the expression below the radical for real numbers be positive will define the function's domain. Since #x^2# will always be positive regardless of the sign of #x#, you need to find the values of #x# that will make #x^2# smaller than #1#, since those are the only values that will make the expression negative. Thus, you must possess #x^2 - 1 >=0# #x^2 >=1# To find, take the square root of each side. #|x| >= 1# Naturally, this indicates that you have #x >= 1" "# and #" "x<=-1# The domain of the function will thus be #(-oo, -1] uu [1, + oo)#. The range of the function will be determined by the fact that the square root of a real number must always be positive. The smallest value the function can take will happen for #x = -1# and for #x=1#, since those values of #x# will make the radical term equal to zero. #sqrt((-1)^2 -1) = 0" "# and #" "sqrt((1)^2 -1 ) = 0# The range of the function will thus be #[0, + oo)#. sqrt(x^2-1) [-10, 10, -5, 5]} graph

Savannah Anderson · Answer

undefined The domain of $ y = \sqrt{x^2 - 1} $ is all real numbers $ x $ such that $ x^2 - 1 \geq 0 $, which means $ x^2 \geq 1 $. So, the domain is $ x $ in all real numbers except $ x = 1 $ and $ x = -1 $. The range of the function is all real numbers $ y $ such that $ y \geq 0 $.