# How do you find the inverse of f(x)=sqrt(x-3)?

May 20, 2017

${f}^{-} 1 \left(x\right) = {x}^{2} + 3$ where $x \ge 0$

#### Explanation:

You can find the inverse function by exchanging $x$ and $y$ and then solve for $y$ in terms of $x$.

$y = \sqrt{x - 3}$ (Original function)
$x = \sqrt{y - 3}$
${x}^{2} = y - 3$
${x}^{2} + 3 = y$
Therefore
${f}^{-} 1 \left(x\right) = {x}^{2} + 3$

The domain of the inverse function, ${f}^{-} 1 \left(x\right)$ is equal to the range of the original function $f \left(x\right)$. The range should be $\left[0 , + \infty\right)$ if the domain of $f \left(x\right)$ is $\left[3 , + \infty\right)$.

May 20, 2017

The inverse is ${f}^{-} 1 \left(x\right) = {x}^{2} + 3$

#### Explanation:

The domain of $f \left(x\right)$ is $x \in \left[3 , + \infty\right)$ and the range is $y \in \left[0 , + \infty\right)$

Let $\textcolor{b l u e}{y} = \sqrt{\textcolor{red}{x} - 3}$

Interchange $x$ and $y$

$\textcolor{red}{x} = \sqrt{\textcolor{b l u e}{y} - 3}$

Now, we express $y$ in terms of $x$

${x}^{2} = y - 3$

$y = {x}^{2} + 3$

Therefore,

${f}^{-} 1 \left(x\right) = {x}^{2} + 3$

The domain and range of ${f}^{-} 1 \left(x\right)$ is the range and domain of $f \left(x\right)$

The domain of ${f}^{-} 1 \left(x\right)$ is $x \in \left[0 , + \infty\right)$ and the range is $y \in \left[3 , + \infty\right)$

Verification

$f \left({f}^{-} 1 \left(x\right)\right) = f \left({x}^{2} + 3\right) = \sqrt{{x}^{2} + 3 - 3} = x$