Answer
$f(x)=x^{2}-4x+5$
$f^{-1}(x)=2+\sqrt{x-1}$
Work Step by Step
$f(x)=x^{2}-4x+5$, for $x\gt2$
Substitute $f(x)$ by $y$:
$y=x^{2}-4x+5$
Group the first two terms on the right side of the equation together:
$y=(x^{2}-4x)+5$
Complete the square for the expression inside parentheses. Do so by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside parentheses and subtracting it from the expression outside of the parentheses. In this case, $b=-4$
$y=\Big[x^{2}-4x+\Big(-\dfrac{4}{2}\Big)^{2}\Big]+5-\Big(-\dfrac{4}{2}\Big)^{2}$
$y=(x^{2}-4x+4)+5-4$
$y=(x^{2}-4x+4)+1$
Factor the expression inside parentheses, which is a perfect square trinomial:
$y=(x-2)^{2}+1$
Take $1$ to the left side and rearrange:
$y-1=(x-2)^{2}$
$(x-2)^{2}=y-1$
Take the square root of both sides:
$\sqrt{(x-2)^{2}}=\sqrt{y-1}$
$x-2=\sqrt{y-1}$
Take $2$ to the right side:
$x=2+\sqrt{y-1}$
Interchange $x$ and $y$:
$y=2+\sqrt{x-1}$
Substitute $y$ by $f^{-1}(x)$:
$f^{-1}(x)=2+\sqrt{x-1}$
The graph of the function and its inverse is shown in the answer section.
