Find the Inverse f(x)=x^2+3 for x>=0

$f (x) = x^{2} + 3$ for $x \geq 0$

Find the range of the given function.

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

$[3, \infty)$

Convert $[3, \infty)$ to an inequality.

$y \geq 3$

Find the inverse.

Interchange the variables.

$x = y^{2} + 3$

Solve for $y$ .

Rewrite the equation as $y^{2} + 3 = x$ .

$y^{2} + 3 = x$

Subtract $3$ from both sides of the equation.

$y^{2} = x - 3$

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

$y = \pm \sqrt{x - 3}$

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.

$y = \sqrt{x - 3}$

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

$y = - \sqrt{x - 3}$

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

$y = \sqrt{x - 3}$

$y = - \sqrt{x - 3}$

$y = \sqrt{x - 3}$

$y = - \sqrt{x - 3}$

$y = \sqrt{x - 3}$

$y = - \sqrt{x - 3}$

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

$f^{- 1} (x) = \sqrt{x - 3}, - \sqrt{x - 3}$

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

$f^{- 1} (x) = \sqrt{x - 3}, x \geq 3$