How to find inverse of $f(x) = x - x^{2}$?

Question

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?

Answer 1

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?

Answer 2

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 $(3/4)$?

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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.

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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.