f inverse of f(a)

Question

$f\left(x\right)=x^{2}-4x+8$, for $x\ge2$

If $a\le0$, find $f^{-1}\left(f\left(a\right)\right)$.

I know $f^{-1}\left(x\right)=2+\sqrt{x-4}$

How to solve my qn then.

Do you know it in the sense that you've proved that $f:[2,\infty)\to \operatorname{im}\left.f\right\rvert_{[2,\infty)}$ is bijective and $f^{-1}(x)=2+\sqrt{x-4}$ for all $x\in \operatorname{im}\left.f\right\rvert_{[2,\infty)}$ ? — Sassatelli Giulio, Mar 31 '24 at 00:45
wth it's just the inverse, u dont need 2 know all that. I want to find the inverse of f(a) moreso. — TheQuantumObsession, Mar 31 '24 at 00:47
If you have trouble writing in English try using https://translate.google.com/ — jjagmath, Mar 31 '24 at 00:56
@TheQuantumObsession Oh, no, I would have to ask other questions and I'm quite done with that. Happy Easter, though. — Sassatelli Giulio, Mar 31 '24 at 00:59
$f(x) = \begin{cases}x^2-4x+8, &\text{for $x\ge 2$}\ \text{what?}&\text{for $x<2$}\end{cases}$ — peterwhy, Mar 31 '24 at 02:15
@TheQuantumObsession You should definitely explain where this problem comes from, at least some context. Otherwise this problem will be closed since it doesn't follow the guideline well. — Liding Yao, Mar 31 '24 at 02:36

score 0 · Answer 1 · answered Mar 31 '24 at 02:08

If $a<0$, since $f(x)$ has domain $[2,\infty)$, $f(a)$ does not exist (or is not defined, depending on your preferred terminology).

The rest of the question is therefore meaningless.

If you would like to check your question and ask it again, I'm sure people would be glad to assist.

score 0 · Answer 2 · answered Mar 31 '24 at 09:39

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

to find the inverse we set $f(x)=y $ and solve for $x$ which you did perfectly

So we have $f^{-1}(x)=2+\sqrt {x-4} , x \in (- \infty,2] $

and $f^{-1}(x)= 2- \sqrt{x-4} , x \in [2,+ \infty)$

In your original question you said $f$ is defined for $x \geq 2$ and $a \leq 0$ so we cannot put $a$ inside of $f$.

Although if we define $f$ for $x \in \mathbb{R}$

now it is meaningful to ask what $f^{-1}(f(a))$ is.

We know that $f^{-1}(x)=y \iff f(y)=x $ from the defenition of inverse functions

so if we sub the $x$ in the first equation with $f(y)$ we get that $f^{-1}(f(y))=y $

if we let $y=a \implies f^{-1}(f(a))=a $ and that's the answer.

2 Answers2