If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$? with the condition : $x > 1$
I solved this problem in this way:
$f(x) = x^2 - 2x + 5 -1 +1 \longrightarrow (x-2)^2 + 1 = f(x) $
$f^{-1}(x) = \sqrt{x-1} + 2$
But I saw someone else solved it in this way:
$f(x) = x^2 - 2x + 1 + 4 \longrightarrow (x+1)^2 + 4 = f(x) $
$f^{-1}(x) = \sqrt{x-4} + 1$
Which one is correct? If the second one is correct why mine is wrong?
After the edit, in the given interval $f(x)$ is bijective and hence invertible. But your method is incorrect because: $$(x-2)^2+1=x^2-4x+5\neq f(x)$$ Your friend is right, but I believe in the second-last step you meant $f(x)=(x-1)^2+4$
$$f(x)=y= x^2-2x+5=x^2-2x+4+1=(x-1)^2+4$$
$$(x-1)^2=y-4$$
$$x-1=\pm\sqrt{y-4} $$
$$x=\pm\sqrt{y-4}+1$$
Interchange $x$ and $y$
$$f^{-1}(x)=y=\pm\sqrt{x-4}+1$$
for $x>1$ inverse of f, $f^{-1}(x)=y=\sqrt{x-4}+1$
