Find the inverse of the function f(x)=x^{2}-8x+4, where

Question

Find the inverse of the function f(x)=x^{2}-8x+4, where <4. State its domain.

anonymous · Answer

$$where    x \le 4$$

anonymous · Answer

@hartnn

hartnn · Answer

can you complete the square ? know how to ?

hartnn · Answer

$f(x)=x^2-8x+4= (x-...)^2-...$

anonymous · Answer

so the answer is inversef(x)=4+sqrt{x+12} . domain is x bigger or equal  -12 ?

hartnn · Answer

to be more precise, 
$f^{-1}(x)=4\pm \sqrt{x+12} $
the domain $x \ge -12 $ is correct.

hartnn · Answer

wait, its given that x<4

hartnn · Answer

so, it'll be only $f^{-1}(x)=4 - \sqrt{x+12}$
with same domain.

anonymous · Answer

how can we understand that it is minus when x<4 ?

hartnn · Answer

4+ anything means >4 
4- anything means <4

hartnn · Answer

does that clear it up ? or you have different doubt altogether ? :P

anonymous · Answer

a little bit :D because it tells about x but we take the function  <4  :/

hartnn · Answer

oh, ....
domain of f(x) means range of its inverse function $f^{-1}(x)$
here, x<4 is domain of f(x), so range of $f^{-1}(x)$ is x<4 , means $f^{-1}(x)$ has least value of 4, so we discard 4+sqrt {x+12} because that will be >4

hartnn · Answer

***range of $f^{-1}(x)$ is $f^{-1}(x)$<4

hartnn · Answer

***maximum value of 4

hartnn · Answer

sorry for the typos.

anonymous · Answer

oh. okey. thank youu :)

hartnn · Answer

welcome ^_^