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

1 Answer
Stefan V. · mason m
Aug 26, 2015

Domain: #(-oo, -1] uu [1, + oo)#
Range: #[0, + oo)#

Explanation:

The domain of the function will be determined by the fact that the expression that's under the radical must be positive for real numbers.

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.

So, you need to have

#x^2 - 1 >=0#

#x^2 >=1#

Take the square root of both sides to get

#|x| >= 1#

This of course means 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)#.

graph{sqrt(x^2-1) [-10, 10, -5, 5]}

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