How to find inverse of $f(x) = x - x^{2}$? Sorry but I don't know how to manipulate it to get all $x$'s on one side, or even with the quadratic, $x - x^{2} - y = 0$.
The question asks why $x$ more-than-or-equal to $1$ is suitable as a domain, which I don't even know where to begin... I don't understand it. I mean do I just enter $2$ into $x$ and solve it?
 A: HINT
The range of the proposed function is given by
$$f(\mathbb{R}) = (-\infty,1/4]$$
But it is not injective.
However, if you restrict its domain to $(-\infty,1/2]$ or $[1/2,+\infty)$, it becomes injective (hence bijective).
Let $f:(-\infty,1/2]\to(-\infty,1/4]$. Hence $f^{-1}$ is denoted by $f^{-1}:(-\infty,1/4]\to(-\infty,1/2]$.
Gathering such considerations, one concludes that its expression is given by:
\begin{align*}
f(f^{-1}(y)) = y & \Longleftrightarrow f^{-1}(y) - [f^{-1}(y)]^{2} = y\\\\
& \Longleftrightarrow [f^{-1}(y)]^{2} - f^{-1}(y) + \frac{1}{4} = \frac{1}{4} - y\\\\
& \Longleftrightarrow \left(f^{-1}(y)  - \frac{1}{2}\right)^{2} = \frac{1}{4} - y\\\\
& \Longleftrightarrow f^{-1}(y) = \frac{1}{2} - \sqrt{\frac{1}{4} - y}
\end{align*}
Based on such discussion, can you handle the other case?
A: Alternative approach:
For the function $f(x)$ to have an inverse function $g(y)$,
this means that :

*

*$g(y) = x \iff f(x) = y.$

*Therefore, for a given value of $y$, you must not have two distinct values $x_1,x_2$ such that $f(x_1) = y = f(x_2).$
So, the question is, how should the domain of $f(x)$ be restricted to prevent this from happening?  To answer that question, you have to explore what the equation $f(x_1) = f(x_2)$ would imply.
$$x_1 - x_1^2 = x_2 - x_2^2 \implies (x_1 - x_2) = x_1^2 - x_2^2 = (x_1 - x_2) \times (x_1 + x_2).$$
Since it is being assumed that $x_1$ and $x_2$ are distinct values, you know that $(x_1 - x_2) \neq 0.$  Therefore, in the equation above, you can divide both sides by the factor of $(x_1 - x_2).$
This implies that
$$1 = (x_1 + x_2).  \tag1 $$
Here, the value of $x = (1/2)$ is acceptable, because then, you would have $x_2 = (1/2) = x_1$, so then, $x_1$ and $x_2$ would not be distinct.
However, if (for example) you specify that the domain of $f(x)$ includes a specific value $~(1/2) + w ~: w > 0$, then the domain must not include the corresponding value $~(1/2) - w.$
So, for example, if the domain of $f(x)$ includes the value $(3/4)$, then it must not include the value $(1/4)$.
For illustration purposes, suppose that the domain of $f(x)$ included both elements of the set $\{ ~1/4, ~3/4\}.$
Then, you have that
$$f(1/4) = (3/16) = f(3/4).$$
In that event, for the inverse function $g(y)$, how would you decide whether $g(3/16)$ is equal to $(1/4)$ or $(3/4)$?

One simple (but somewhat arbitrary) specification for the domain of $x$, based on the potential conflict detailed in (1) above, would be $(1/2) \leq x.$
An alternative specification would be $(1/2) \geq x.$
These specifications are simple and arbitrary.  You could have more complicated specifications.  For example, you could specify the domain of
$$\{ ~x ~: ~x < 0 ~\} \cup \{ ~x ~: ~1/2 \leq x \leq 1 ~\}. \tag2 $$
Specifying the disjoint union, shown in (2) above, while somewhat convoluted, would still prevent the conflict detailed by the equation in (1) above.

To answer the specific question, why would the set $\{ ~x ~: ~1 \geq x ~\} ~$ be suitable as a domain: the answer is because if the domain is restricted to this set, then the conflict detailed by the equation in (1) above would then be prevented from happening.
