The Inverse of the Square Root Function

1. Switch $\color{#ec0000}{x}$ and $\color{#333fb1}{y}$ in the function rule.

2. Isolate $\color{#333fb1}{y}.$

3. Calculate the domain restriction of the inverse from the function’s range. For $f(x)=a\sqrt{b(x-h)}+k:$
   $\text{dom}\ f^{-1}=\text{ran}\ f=\ ]-\infty,k]$ if $a<0$
   or
   $\text{dom}\ f^{-1}=\text{ran}\ f=\ [k,+\infty[$ if $a>0.$

4. Write the rule of the inverse.

Find the rule of the inverse of the function $f(x)=2\sqrt{x-5}+10.$

1. Switch $x$ and $y$ in the rule
   $$\begin{align}\color{#333fb1}{y}&=2\sqrt{\color{#ec0000}{x}-5}+10\\ \color{#ec0000}{x}&=2\sqrt{\color{#333fb1}{y}-5}+10\end{align}$$

2. Isolate $y$
   $$\begin{align}x&=2\sqrt{y-5}+10\\ x-10&=2\sqrt{y-5}\\ \dfrac{x-10}{2}&=\sqrt{y-5}\\ \color{#ec0000}{\left(\color{black}{\dfrac{x-10}{2}}\right)^2}&=\color{#ec0000}{\left(\color{black}{\sqrt{y-5}}\ \right)^2}\\ \dfrac{(x-10)^2}{4}&=y-5\\ \dfrac{1}{4}(x-10)^2+5&=y\end{align}$$

3. Calculate the restriction on the inverse’s domain
   For $f(x),$ $a=2$ and $k=10.$
   
   The parameter $a$ of $f(x)$ is positive, suggesting the function is defined above its vertex. So, the inverse function’s domain $f^{-1}(x),$ which is equivalent to the range of $f(x),$ is the following.
   $$\begin{align}\text{dom}\ f^{-1}=\text{ran}\ f&=[k,+\infty[\\&=[10,+\infty[\end{align}$$

4. Give the rule
   The function’s inverse $f(x)$ is $f^{-1}(x)=\dfrac{1}{4}(x-10)^2+5$ where $x\ge10.$

The curves of $\color{#333fb1}{f(x)}$ and of $\color{#ec0000}{f^{-1}(x)}$ from the previous example are found on the following Cartesian plane.

It is possible to plot the inverse of a function by interchanging the coordinates $x$ and $y$ of certain points. For example, the vertex $(\color{#333fb1}{5},\color{#333fb1}{10})$ becomes the vertex $(\color{#ec0000}{10},\color{#ec0000}{5})$ and the point $(\color{#333fb1}{9},\color{#333fb1}{14})$ becomes $(\color{#ec0000}{14},\color{#ec0000}{9}).$

It can also be said that $\color{#ec0000}{f^{-1}(x)}$ corresponds to the reflection of $\color{#333fb1}{f(x)}$ regarding the line $y=x.$ It is therefore possible to graph the inverse by reflection. For it to work, the scale of the $x$- and $y$-axes must have a ratio of $1:1.$