Answer
The inverse function for$f\left( x \right)=\sqrt{x}+1$ is ${{f}^{-1}}\left( x \right)={{\left( x-1 \right)}^{2}},x\ge 1$ .
Work Step by Step
Consider the provided function:
$f\left( x \right)=\sqrt{x}+1$
Let $y=\sqrt{x}+1$
The steps to find the inverse function $y=f\left( x \right)$ are as follows:
Step1: Interchange $x$ and $y$.
$x=\sqrt{y}+1$
Step 2: Solve the above equation for $y$ .
$x-1=\sqrt{y}$
Square both sides of the above equation,
Thus,
$y={{\left( x-1 \right)}^{2}}$
Step 3: Replace $y$ with ${{f}^{-1}}\left( x \right)$ .
${{f}^{-1}}\left( x \right)={{\left( x-1 \right)}^{2}},x\ge 1$
Therefore, the inverse function for $f\left( x \right)=\sqrt{x}+1$ is ${{f}^{-1}}\left( x \right)={{\left( x-1 \right)}^{2}},x\ge 1$.
