Let $f:\mathbb{R}\to\mathbb{R}\space \space \space f(x)=\left\{\begin{array}{lc}x^2-2x,&x\leq0\\-x^2-2x,&x>0\end{array}\right.$. Find $(f^{-1})'(3)$.
My attempt:
First, it's easy to see that $f(x)$ is injective $\forall x \in \mathbb{R}.$ By doing the first derivative table we see that $Im_f=\mathbb{R}.$ so then $f$ is surjective and then $f$ is bijective and it has an inverse function.
$$f^{-1}(f(x))=x$$
differentiating both sides we get that $(f^{-1}(f(x)))'=\frac 1{f'(x)}$.
$f(x)=3\implies x = -1$ unique solution so then $\boxed{(f^{-1})'(3)=\frac 1{f'(x)}=-\frac 14}$
Second solution: (which i don't really understand what happens here)
I want to define $f^{-1}(x).$
First case ($x\leq0):$
$\implies f(x)=x^2-2x \space \space \space \forall x\leq0.$
$$f^{-1}(x)=\pm\sqrt{x+1}+1 \space\space\space \forall x\geq -1.$$
so then: $f^{-1}(x)=\pm\sqrt{x+1}+1 \space \space \space \forall \space x\in[-1,0]?$ Also when do we use the $+$ sign and the $-$ for the square root?
Second case ($x>0$):
$\implies f(x)=-(x+1)^2+1\implies f^{-1}(x)=\pm\sqrt{1-x}-1 \space \space \space \forall x\in(0,1]?$ same question about the sign of the square root.
so then we get
$$\boxed{f^{-1}(x)=\left\{\begin{array}{lc}\pm\sqrt{x+1}+1,&x\in\lbrack-1,0\rbrack\\\pm\sqrt{1-x}-1,&x\in(0,1\rbrack\end{array}\right.}$$
But by differentiating is we can't find $(f^{-1})'(3)$. But $f^{-1}(x)$ should be defined on all $\mathbb{R}$..
So... What I am doing wrong? What is really happening here?
Also when I try to graph them:
We can see that $f^{-1}(x)$ takes values only in $[-1,1]$.
The difficulty seems to be writing down $f^{-1}(x)$.
To solve this question, we look separately at the range of $f$ , for it is on that range that $f^{-1}$ can be defined.
Let us look at $x^2 - 2x$ for $x \leq 0$. As $x$ decreases, this goes to $+ \infty$. The mimimum of this function is at $x = 0$, with the value $0$. So this piece of the function has range $[0,\infty)$.
Next, $-x^2-2x$ for $x > 0$. Clearly, as $x \to \infty$, this goes to $-\infty$. Furthermore, this is equal to $-x(x+2)$, and as $x \to 0$ this approaches zero. Hence,the range of this function is exactly $(-\infty,0)$.
So $f$ is surjective. It is also injective, as seen by you.
Now, given $y \in \mathbb R$, suppose $y \geq 0$. Then $y$ is in the range of the first piece, so there is $x \leq 0$ so that $x^2 - 2x = y$. Completing the square and solving, $x = 1 \pm \sqrt{y+1}$, but since $x \leq 0$ we must have $-$, not $+$ in the answer. Therefore, $x = 1 - \sqrt{y+1}$, so $f^{-1}(y) = 1 - \sqrt{y+1}$.
Similarly, if $y < 0$, then $y$ is in the range of the second piece, so there is $x \geq 0$ such that $-x^2 - 2x = y$. Solving, $x = -1 \pm \sqrt{1-y}$, but we want $x \geq 0$ so the sign is $+$ not $-$, hence $f^{-1}(y) = -1 + \sqrt{1-y}$.
Consequently, we have : $$ f^{-1}(y) = \begin{cases} 1 - \sqrt{y+1} \quad y \geq 0 \\ \sqrt{1-y} - 1 \quad y < 0 \end{cases} $$
Since $3$ belongs to the first piece, $f^{-1}$ is differentiable at $3$, and the derivative is $\frac {-1}{2 \sqrt{y+1}} = \frac {-1}4$ at the point $3$.
Your mistake was not knowing which sign, $\pm$ to choose from the two quadratic solutions.
To get the meaning of the green graph, note that if $x = y^2 - 2y$ for some real number $y$ then $x +1 = y^2 - 2y + 1 \geq 0$ so $x \geq -1$. This is why the graph does not go behind $x = -1$ : you do not have real solutions beyong that point. Since you have already restricted $x \leq 0$, the green graph is confined to $-1 \leq x \leq 0$. Something similar happens with the yellow graph.
Neither of these is an inverse anyway : $x = y^2 - 2y$ is not the graph of the inverse of the function $y = x^2 - 2x$. This is a misconception.
You may graph $f^{-1}$ to see the correctness of the answer.