# Thread: inverse of a function

1. ## inverse of a function

consider the function f(x)=(x-1)^2+2

1. Determine the inverse of f(x)

2. Determine the domain and range of f(x) and its inverse f^-1(x)

As for the range, since $f(x)=(x-1)^2+2\implies\sqrt{f(x)-2}=|x-1|,$ hence, we require that $f(x)\ge2,$ thus the range is $y\ge2.$ (The domain is all reals, of course.)

3. Originally Posted by william
consider the function f(x)=(x-1)^2+2

1. Determine the inverse of f(x)

2. Determine the domain and range of f(x) and its inverse f^-1(x)
$y=(x-1)^2+2$

for finding inverse, interchange x and y and solve for y,

$x=(y-1)^2+2$

$x-2 = (y-1)^2$

$\pm \sqrt{x-2}=y-1$

$y=1\pm \sqrt{x-2}$

$f^{-1}(x)=1\pm \sqrt{x-2}$

For f(x)

Domain of f(x) $= \{x\in \mathbb{R}\}$

Range of f(x) $= \{y\in \mathbb{R}\;| \;y\ge2\}$

Now, For $f^{-1}(x)$

Domain of $f^{-1}(x) =\{x\in \mathbb{R}\;|\;x\ge 2\}$

Range of $f^{-1}(x) = \{y\in \mathbb{R}\}$

4. Strictly speaking $f(x)= (x-1)^2+ 2$ does NOT have an inverse because it is not a "one to one" function.

Yes, you can recognize that the graph of y= f(x) has a vertex at (1, 2) and then break it into TWO functions:
$g(x)= (x-1)^2+ 2$ with domain $1\le x$ which then has range $2\le y$ and write $g^{-1}(x)= 1+ \sqrt{x-2}$ or
$h(x)= (x-1)^2+ 2$ with domain $x\le 1$ which also has range $2\le y$ and write $h^{-1}(x)= 1- \sqt{x- 2}$.

But neither of those functions is f(x) so, strictly speaking, neither of those is the inverse of f.