Q.Let $A = \{-1, 0, 1, 2\}$, $B = \{-4, -2, 0, 2\}$ and $f, g: A \to B$ be functions defined by $f(x) = x^2 - x$, $x \in A$ and $g(x) = 2\left|x - \frac{1}{2}\right| - 1$, $x \in A$. Are $f$ and $g$ equal? Justify your answer. (Hint: One may note that two functions $f: A \to B$ and $g: A \to B$ such that $f(a) = g(a) \ \forall a \in A$, are called equal functions).

Concept: two functions $f,g:A\to B$ are equal when they share domain and codomain and $f(x)=g(x)$ for every $x\in A$. Here both have domain $A=\{-1,0,1,2\}$ and codomain $B=\{-4,-2,0,2\}$, so we just compare outputs.

Compute $f(x)=x^2-x$:

• $f(-1)=1+1=2$
• $f(0)=0$
• $f(1)=1-1=0$
• $f(2)=4-2=2$

Compute $g(x)=2\left|x-\tfrac12\right|-1$:

• $g(-1)=2\cdot\tfrac32-1=2$
• $g(0)=2\cdot\tfrac12-1=0$ …

What "equal functions" means

Two functions $f,g:A\to B$ are equal precisely when they agree at every input: $f(a)=g(a)$ for all $a\in A$ (with the same domain and codomain, which both share here). So the task is just a point-by-point comparison over $A=\{-1,0,1,2\}$.

Evaluate $f(x)=x^2-x$

1. $f(-1)=(-1)^2-(-1)=1+1=2$
2. $f(0)=0^2-0=0$
3. $f(1)=1^2-1=0$
4. $f(2)=2^2-2=2$

So $f$ gives $(2,\,0,\,0,\,2)$ on $(-1,0,1,2)$.

Evaluate $g(x)=2\left|x-\tfrac12\right|-1$

1. $g(-1)=2\left|-\tfrac32\right|-1=2\cdot\tfrac32-1=3-1=2$
2. $g(0)=2\left|-\tfrac12\right|-1=2\cdot\tfrac12-1=1-1=0$
3. $g(1)=2\left|\tfrac12\right|-1=2\cdot\tfrac12-1=1-1=0$
4. $g(2)=2\left|\tfrac32\right|-1=2\cdot\tfrac32-1=3-1=2$

So $g$ also gives $(2,\,0,\,0,\,2)$ on $(-1,0,1,2)$.

Compare

Method: Testing whether two functions are equal

Use this whenever a problem gives you two functions and asks "are they equal?" — the two formulas will usually look different on purpose.

Steps

Step 1: Check they share the same domain and codomain

Two functions $f$ and $g$ can only be equal if they are declared over the same domain and codomain. If the domains differ, stop immediately — they are unequal, no matter how similar the formulas look.

Step 2: Compare outputs on every element of the domain

Equality of functions means $f(a)=g(a)$ for every $a$ in the common domain — not that the two algebraic expressions are identical for all real numbers. When the domain is a small finite set, evaluate both functions at each element and tabulate.

Step 3: Decide …

Common Mistakes

Mistake 1: Concluding $f\ne g$ because the two formulas are algebraically different

Why it's wrong: $x^2-x$ and $2\left|x-\tfrac12\right|-1$ genuinely differ as expressions on $\mathbb{R}$ (at $x=3$ they give $6$ and $4$), so a student who compares formulas concludes "not equal". Correct approach: function equality only requires agreement on the given domain $A=\{-1,0,1,2\}$; evaluate both at each of those four points and compare outputs, not the general formulas.

Mistake 2: Mis-evaluating the modulus in $g$ …