Find the Inverse f(x)=x^2-1 , x>=0

$f (x) = x^{2} - 1$ , $x \geq 0$

Step 1

Find the range of the given function.

The range is the set of all valid $y$ values. Use the graph to find the range.

$[- 1, \infty)$

Convert $[- 1, \infty)$ to an inequality.

$y \geq - 1$

Step 2

Find the inverse.

Interchange the variables.

$x = y^{2} - 1$

Solve for $y$ .

Rewrite the equation as $y^{2} - 1 = x$ .

$y^{2} - 1 = x$

Add $1$ to both sides of the equation.

$y^{2} = x + 1$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{x + 1}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{x + 1}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{x + 1}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{x + 1}$

$y = - \sqrt{x + 1}$

$y = \sqrt{x + 1}$

$y = - \sqrt{x + 1}$

$y = \sqrt{x + 1}$

$y = - \sqrt{x + 1}$

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \sqrt{x + 1}, - \sqrt{x + 1}$

Step 3

Find the inverse using the domain and the range of the original function.

$f^{- 1} (x) = \sqrt{x + 1}, x \geq - 1$

Step 4