Answer
$f(x)=x^{2}-2x+6$
$f^{-1}(x)=1+\sqrt{x-5}$
Work Step by Step
$f(x)=x^{2}-2x+6$, for $x\ge1$
Substitute $f(x)$ by $y$:
$y=x^{2}-2x+6$
Solve the equation for $x$. Start by completing the square on the right side.
$y=(x^{2}-2x)+6$
Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside the parentheses and subtracting it outside of it. In this case, $b=-2$
$y=\Big[x^{2}-2x+\Big(-\dfrac{2}{2}\Big)^{2}\Big]+6-\Big(-\dfrac{2}{2}\Big)^{2}$
$y=(x^{2}-2x+1)+6-1$
$y=(x^{2}-2x+1)+5$
Factor the expression inside the parentheses, which is a perfect square trinomial:
$y=(x-1)^{2}+5$
Take $5$ to the left side:
$y-5=(x-1)^{2}$
Take the square root of both sides:
$\sqrt{y-5}=\sqrt{(x-1)^{2}}$
$\sqrt{y-5}=x-1$
Take $1$ to the left side and rearrange the equation:
$x=1+\sqrt{y-5}$
Interchange $x$ and $y$:
$y=1+\sqrt{x-5}$
Substitute $y$ by $f^{-1}(x)$:
$f^{-1}(x)=1+\sqrt{x-5}$
The graph of both the original function and its inverse is shown in the answer section.
