Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, $x < 0, a = 1$ The textbook answer is $\frac {1}{3}$, I went through all the steps, but couldn't interpret it.
Below were my steps.
Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, x < 0, a = 1
First I tried to find the inverse, rearrange the equation to $$x = y - \frac{2}{y} \implies x = \frac{y^{2}-2}{y}$$
Rearrange, $xy=y^{2}-2 \implies y^{2}-xy-2 = 0$
Then, I tried to use the equation $\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$, and I got:
$y = \frac{x \pm \sqrt{x^{2}+8}}{2}=f^{-1}$
Next, I tried to find $(f^{-1}){'}$, for both $+$ and $-$
$\frac{1}{2} \cdot (x' + ((x^{2}+8)^{\frac{1}{2}})')$
Apply chain rule:
$((x^{2}+8)^{\frac{1}{2}})'$ = $\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x$
Then do the same for -
And I got $(f^{-1})'$ = $\begin{cases} \frac{1}{2}\cdot(1+\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \\ \frac{1}{2}\cdot(1-\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \end{cases}$
When a = 1,
$(f^{-1})'=\begin{cases} \frac{\sqrt{9}+1}{2\sqrt{9}} = \frac{2}{3} \\ \frac{\sqrt{9}-1}{2\sqrt{9}} = \frac{1}{3} \end{cases}$
I need some help. It says when $x < 0$, but both of my answers are positive. How do I interpret it?
 A: The domain of $f$ must be the range of $f^{-1}$, i.e. $f^{-1}$ must only output negative values for any value of $x$. So, you should reject the solution $y=\frac{x+\sqrt{x^2+8}}{2}$ and notice $$\frac{x-\sqrt{x^2+8}}{2} \lt \frac{x-\sqrt{x^2}}{2} \le 0 $$ which satisfies our requirement and the corresponding answer is $\frac 13$.
A: So the issue is you have not used $x<0$. By the way be careful when you use a variable like x. in one part you have used it for the domain and in one for the range set.
By using the quadratic formula you have obtained two solutions, so your question is which is the right one at $a=1$. When $a=1$
$$ 1= x-\frac{2}{x} $$
Hence,
$$x^2-x-2=0 $$
using the quadratic formula you will get two solution +1, -2, so you have only use the negative one. And hence your answer will be 1/3.
Another interesting way to solve this is, as you $f(-2)=1=a$, differentiate $f^{-1}(f(x))=x$ at x=-2 using the chain rule and you should get your required.
Hope this helped.
A: The domain of $f$ is the interval $(-\infty,0)$.
Thus, $g:=f^{-1}$ should be such that its range is $(-\infty,0)$.
This way, one can see that
$$
g(s) = \frac{s-\sqrt{s^2+8}}{2}
$$
is the proper solution for the inverse of $f$,
(because we always have $g(s)<0$), but
$g(s) = \frac{s+\sqrt{s^2+8}}{2}$ would not be, because its values
can be positive (for example we have $g(0)>0$).
Now that the proper inverse function is known, we can see from your calculations
that $\frac13$ is the correct answer and $\frac23$ is wrong.

It says when $x < 0$, but both of my answers are positive. How do I interpret it?

While the output of the function $g=f^{-1}$ should be negative,
its derivative $g'=(f^{-1})'$ can be positive, because these are different objects.
