Find the inverse of f(x) = x\(\small\mathsf{ ^2 }\) − 1 then click the Answer button below.
First, rewrite f(x) to y:
y = x\(\small\mathsf{ ^2 }\) - 1.
Next, interchange the y and x variables:
x = y\(\small\mathsf{ ^2 }\) - 1.
Next, solve for y.
| x + 1 = y\(\small\mathsf{ ^2 }\) |
| \(\small\mathsf{ \sqrt{x+1} = \sqrt{y^{2}} }\) |
| \(\small\mathsf{ \pm \sqrt{x + 1} = y }\) |
| \(\small\mathsf{ f^{-1}(x) = \pm \sqrt{x+1} }\) |
Note that there are two cases: \(\small\mathsf{ f^{-1}(x) = +\sqrt{x + 1} }\) and \(\small\mathsf{ f^{-1}(x) = -\sqrt{x + 1} }\).
But we are dealing with a square root and remember that (x + 1) is not negative to have an answer that is a real number. This rule applies to all even roots (square root, fourth root...). We need to set this quantity so (x + 1) > 0.
| x + 1 > 0 | Subtract 1 from both sides |
| x > -1 |
