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.

Answer 1

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.

Answer 2

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

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.