Question

given $f(x) = x^2 + 3$ for $x \geq 0$, find $f^{-1}(x)$. $f^{-1}(x) = $

Explanation:

Step1: Set $y=f(x)$

$y = x^2 + 3$

Step2: Swap $x$ and $y$

$x = y^2 + 3$

Step3: Solve for $y$

$y^2 = x - 3$
Since $x \geq 0$ for the original function, the range of $f(x)$ is $y \geq 3$, so for the inverse, $x \geq 3$, and we take the positive root:
$y = \sqrt{x - 3}$

Step4: Replace $y$ with $f^{-1}(x)$

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

Answer:

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