Miscellaneous · Q5: Are $x^2-x$ and $2\left|x-\tfrac12\right|-1$ equal?

Basics you need

Solution

1. What we must check

   $f$ and $g$ have the same domain $A$ and the same codomain $B$⁠. By the hint, they are equal exactly when

   $$\begin{aligned} f(-1) &= g(-1), & f(0) &= g(0),\\ f(1) &= g(1), & f(2) &= g(2). \end{aligned}$$

   If even one of these failed, $f$ and $g$ would not be equal.

   "Equal functions" is a "for all $a \in A$" statement. Because $A$ has exactly four elements, four comparisons cover every case (Basic 4).
   A function $A \to B$ is determined by its value at each element of $A$⁠. Basics: what a function is

2. Compute $f(x) = x^2 - x$ on $A$

   $$\begin{aligned} f(-1) &= (-1)^2 - (-1) = 1 + 1 = 2,\\ f(0) &= 0^2 - 0 = 0,\\ f(1) &= 1^2 - 1 = 0,\\ f(2) &= 2^2 - 2 = 4 - 2 = 2. \end{aligned}$$
   Substitute each element of $A$ in turn. Brackets around $-1$ matter: $(-1)^2 = 1$⁠, and subtracting $-1$ means adding 1. All four values, 2, 0, 0, 2, lie in $B$⁠, so $f$ really is a function from $A$ to $B$⁠.
   $(-a)^2 = a^2$ and $-(-a) = a$⁠.

3. Compute $g(x) = 2\left|x - \frac12\right| - 1$ on $A$

   First the distances from $\frac12$⁠:

   $$\begin{aligned} \left|-1 - \tfrac12\right| &= \left|-\tfrac32\right| = \tfrac32,\\ \left|0 - \tfrac12\right| &= \left|-\tfrac12\right| = \tfrac12,\\ \left|1 - \tfrac12\right| &= \tfrac12,\\ \left|2 - \tfrac12\right| &= \tfrac32. \end{aligned}$$

   Then double and subtract 1:

   $$\begin{aligned} g(-1) &= 2\cdot\tfrac32 - 1 = 3 - 1 = 2,\\ g(0) &= 2\cdot\tfrac12 - 1 = 1 - 1 = 0,\\ g(1) &= 2\cdot\tfrac12 - 1 = 1 - 1 = 0,\\ g(2) &= 2\cdot\tfrac32 - 1 = 3 - 1 = 2. \end{aligned}$$
   Working out the absolute values first, separately, avoids the usual slip of dropping the bars and getting $2(-\frac32) - 1 = -4$ for $g(-1)$⁠. The four values 2, 0, 0, 2 again lie in $B$⁠, so $g$ is also a genuine function from $A$ to $B$⁠.
   $|t| = -t$ when $t < 0$⁠, so $\left|-\frac32\right| = \frac32$ and $\left|-\frac12\right| = \frac12$ (Basic 2).

4. Compare, value by value

   $a$	$-1$	$0$	$1$	$2$
   $f(a) = a^2 - a$	2	0	0	2
   $\left|a - \frac12\right|$	$\frac32$	$\frac12$	$\frac12$	$\frac32$
   $g(a)$	2	0	0	2
   equal?	yes	yes	yes	yes

   $f(a) = g(a)$ for every $a \in A$⁠. Since $f$ and $g$ also share the domain $A$ and codomain $B$⁠, $f = g$.

   All four required equalities from Step 1 hold, and $A$ has no other elements, so the definition of equal functions is satisfied completely.
   Equal functions: same domain, same codomain, same value at every point (Basic 1).

5. Why the answer depends on the domain

   The two formulas are not the same on the whole real line. At $x = \frac12$ (not in $A$⁠), $x^2 - x = -\frac14$ but $2\left|x-\frac12\right| - 1 = -1$⁠. So where do the formulas agree? Solve $x^2 - x = 2\left|x - \frac12\right| - 1$ over $\R$⁠, splitting on the sign inside the bars:

   If $x \ge \frac12$⁠: $\left|x-\frac12\right| = x - \frac12$⁠, so the right side is $2x - 2$⁠, and

   $$\begin{aligned} & x^2 - x = 2x - 2\\ \iff\ & x^2 - 3x + 2 = 0\\ \iff\ & (x-1)(x-2) = 0\\ \iff\ & x = 1 \text{ or } x = 2. \end{aligned}$$

   If $x < \frac12$⁠: $\left|x-\frac12\right| = \frac12 - x$⁠, so the right side is $-2x$⁠, and

   $$\begin{aligned} & x^2 - x = -2x\\ \iff\ & x^2 + x = 0\\ \iff\ & x(x+1) = 0\\ \iff\ & x = 0 \text{ or } x = -1. \end{aligned}$$

   All four roots lie in the case they were found in ($1, 2 \ge \frac12$ and $0, -1 < \frac12$⁠). So the formulas agree at exactly the four real numbers $-1, 0, 1, 2$⁠, which is precisely the set $A$⁠.

   This explains the question: $A$ was chosen as exactly the set where the formulas agree. If the domain were $\R$⁠, or even $\{-1, 0, \frac12, 1, 2\}$ (with a codomain such as $\R$ large enough to hold the new values), the functions would not be equal. Equality of functions is always relative to the domain.
   Absolute value split: $|t| = t$ for $t \ge 0$ and $-t$ for $t < 0$⁠; a product is zero exactly when a factor is zero.

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