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

1. anewbie

$where x \le 4$

2. anewbie

@hartnn

3. hartnn

can you complete the square ? know how to ?

4. hartnn

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

5. anewbie

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

6. hartnn

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

7. hartnn

wait, its given that x<4

8. hartnn

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

9. anewbie

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

10. hartnn

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

11. hartnn

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

12. anewbie

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

13. hartnn

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

14. hartnn

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

15. hartnn

***maximum value of 4

16. hartnn

sorry for the typos.

17. anewbie

oh. okey. thank youu :)

18. hartnn

welcome ^_^