ANSWER
[tex] {f}^{ - 1} =\pm \sqrt{x + 1} [/tex]
EXPLANATION
The given function is
[tex]f(x) = {x}^{2} - 1[/tex]
Let
[tex]y = {x}^{2} - 1[/tex]
Interchange x and y to get:
[tex]x= {y}^{2} - 1[/tex]
Solve for y,
[tex]x + 1 = {y}^{2} [/tex]
Take square root of both sides to get,
[tex] \pm \sqrt{x + 1} = y[/tex]
This is the same as:
[tex]y = \pm \sqrt{x + 1} [/tex]
Therefore the inverse is
[tex] {f}^{ - 1} =\pm \sqrt{x + 1} [/tex]
