What is the best way to find $g(x) = f^{-1}(x)$?

Question

so the problem I have is if $f(x) = \sqrt{x+3} - 2$ and it asks to find the solution of $f(x) = f^{-1}(x)$. So i know to find the inverse, which I got as $f^{-1}(x) = (x+2)^2-3$. So to find the solution, (I can use a calculator), should I graph both of them and find the intersection(s), or should I actually algebraically solve and set the answer to 0 and graph that polynomial to get the solutions, or is there a better way?

Thanks! ~ Nathan

Answer 1

Denoting f{x) =y, just interchange x and y . It is that simple and also it is done!

$y=\sqrt{x+3}-2$ becomes

$x=\sqrt{y+3}-2, $ or $ y = (x+2)^2 -3. $

You can graph $ f(x) $ and $ f^{-1}(x) $ just to check that they indeed intersect and reflect on the line $x=y,$ and nothing more needs to be done.

Answer 2

One possibility is to note that $f(x) = f^{-1}(x)$ is equivalent to (minus problems with domains!) having $f(f(x)) = x$, or even $f^{-1}(f^{-1}(x))=x$, so if one of $f(f(x))$ or $f^{-1}(f^{-1}(x))$ is particularly nice that may be the most efficient way. Also, fixed points, i.e. points with $f(x) = x$ (or $f^{-1}(x) = x$) have $f(x) = f^{-1}(x)$.

So in your example, it may be easiest to solve the quadratic equation $f^{-1}(x) = x$, get two solutions, then solve the quartic $f^{-1}(f^{-1}(x))=x$ with the aid of already knowing two solutions turning it into just solving another quadratic.

Answer 3

You need to solve equation $$\sqrt{x+3}-2=(x+2)^2-3$$ $$\sqrt{x+3}=(x+2)^2-1=(x+3)^2-2(x+3)$$ Taking $x+3=t^2$ we can write

$$t^4-2t^2-t=0$$ this equation after factoring turn to $$t(t+1)(t^2-t-1)=0$$ $$t_1=0,t_2=-1,t_3=\frac{1+\sqrt5}{2},t_4=\frac{1-\sqrt5}{2}$$ $$x_i=t_i^2-3,i=1,2,3,4$$