Exercises — Concept of a function — input, output, mapping

The golden rule has two halves: every input must fire (no input left blank) and each input fires exactly once (no input with two arrows).

• (i) Every input (1,2,3) has exactly one arrow. ✓ Function.
• (ii) Input $2$ sends arrows to both $b$ and $c$ — two outputs for one input. ✗ Not a function. (Look at the two arrows leaving "2" in the middle panel.)
• (iii) Input $2$ has no arrow at all — it is left blank. ✗ Not a function, because the domain $A$ contains $2$ and every element of the domain must map somewhere.

Answer: only (i).

"Apply the rule" means replace every $x$ by the input, then simplify.

• $f(0)=3(0)-4=-4$
• $f(2)=3(2)-4=6-4=2$
• $f(-1)=3(-1)-4=-3-4=-7$

Answers: $-4,2,-7$.

• $f(3)=3^2+1=10$
• $f(-3)=(-3{)}^2+1=9+1=10$ — different input, same output. Allowed! The rule only forbids one input giving two outputs, not two inputs sharing an output.
• $f(a+1)$: replace the whole $x$ by $a+1$: $f(a+1)=(a+1{)}^2+1=a^2+2a+1+1=a^2+2a+2.$

Answers: $10,10,a^2+2a+2$.

A piecewise rule is one machine with a switch: check which condition your input satisfies first, then use only that branch.

• $g(4)$: $4\ge 0$, top branch $\Rightarrow 2(4)=8$.
• $g(-4)$: $-4<0$, bottom branch $\Rightarrow -(-4)=4$.
• $g(0)$: $0\ge 0$ (the $=$ is in the top branch), so top $\Rightarrow 2(0)=0$.

Notice $0$ appears in exactly one branch — that is on purpose, so $g$ stays single-valued.

Answers: $8,4,0$.

The only thing forbidden is dividing by zero. The denominator $x-2$ is zero when $x=2$. Every other real number is fine.

Domain $=\mathbb{R}\setminus \{2\}$ (all reals except $2$).

We ask: what values can come out? Complete the square to see the lowest point. $x^2-4x+7=(x^2-4x+4)+3=(x-2{)}^2+3.$ A square $(x-2{)}^2$ is always $\ge 0$, and it hits $0$ when $x=2$. So the smallest output is $0+3=3$, and outputs grow without bound above that.

Range $=[3,\infty )$. (Domain is all of $\mathbb{R}$ — don't confuse the two: you may input any real, but outputs start at 3.)

A vertical line $x=c$ collects all points sharing that input $c$. If it crosses the curve more than once, that single input $c$ has several outputs — no function.

Draw $x=3$: it meets the circle at $(3,4)$ and $(3,-4)$. Two outputs for input $3$. ✗ Not a function.

Repair: keep only the top half, $y=\sqrt{25-x^2}$. Now every vertical line for $-5\le x\le 5$ hits exactly one point. ✓ Function, with domain $[-5,5]$ and range $[0,5]$.

A real square root needs its inside $\ge 0$ (negatives have no real root). So require $6-2x\ge 0\Rightarrow 6\ge 2x\Rightarrow x\le 3.$ Domain $=(-\infty ,3]$.

Solving $y^2=x$ for $y$ gives $y=\pm \sqrt{x}$ — the $\pm$ is exactly the "two outputs" that ruins single-valuedness. Split the sign:

• $f_1(x)=+\sqrt{x}$, domain $[0,\infty )$ (need $x\ge 0$), range $[0,\infty )$.
• $f_2(x)=-\sqrt{x}$, domain $[0,\infty )$, range $(-\infty ,0]$.

Each is single-valued on its own; together their graphs recreate the whole sideways parabola. Check: at $x=4$, $f_1(4)=2$ and $f_2(4)=-2$ — the two branches.

Distance is never negative, and "distance from 10" is captured by absolute value: $p(x)=|x-10|.$

• $p(3)=|3-10|=|-7|=7$
• $p(15)=|15-10|=|5|=5$

Since a distance is $\ge 0$ and can be as small as $0$ (at $x=10$) or arbitrarily large, range $=[0,\infty )$.

"Meet at $x=1$" means both branches give the same output as we approach $x=1$. The top branch owns $x=1$: value $1+c$. The bottom branch approaches $3(1)-2=1$. Set equal: $1+c=1\Rightarrow c=0.$ Answer: $c=0$. With $c=0$ the machine has no sudden jump at $x=1$.

Sharing an output is fine: take $f(x)=x^2$. Then $f(3)=9=f(-3)$ with $3\ne -3$. The definition never says outputs must be unique to their input.

Two outputs for one input is impossible by definition: suppose $f(a)=y_1$ and $f(a)=y_2$ with $y_1\ne y_2$. The definition of a function says $f$ assigns to each input exactly one value. "Exactly one" and "$y_1\ne y_2$ both assigned" directly contradict each other. Hence no such function exists — the very word function forbids it.

This asymmetry is the heart of the concept: many-to-one is allowed, one-to-many is banned.

• Function? Yes. Every input has exactly one output — the output just never changes. "Same output every time" is more deterministic, not less.
• Domain: $\mathbb{R}$ (you may input any real).
• Range: $\{7\}$ — a single-element set, because only 7 ever comes out.
• Vertical line test: the graph is a horizontal line at height 7. Any vertical line meets it exactly once. ✓

Degenerate but perfectly valid.

The rule "every input has exactly one output" must hold for every element of the domain. If there are no inputs, there is nothing that could break the rule — no input exists to have zero or two outputs. So the condition is satisfied vacuously (true because there is nothing to check).

Yes, it is a (trivial) function, called the empty function. This is the limiting case at the "no inputs" end and shows the definition is watertight even when pushed to emptiness.

Each input must choose exactly one output. Input $1$ has $2$ choices ($u$ or $v$); independently input $2$ has $2$ choices; input $3$ has $2$ choices. Multiply the independent choices: $2\times 2\times 2=2^3=8.$ In general, from a domain of size $m$ to a codomain of size $n$ there are $n^m$ functions.

Answer: $8$.

A function forbids one input mapping to two outputs but allows two inputs mapping to one output. The domain of $\sqrt{6-2x}$ is ::: $(-\infty ,3]$ The range of $x^2-4x+7$ is ::: $[3,\infty )$ Number of functions from a 3-set to a 2-set ::: $2^3=8$