Question bank — Domain, codomain, range

False — the codomain is declared by us, so the same rule $f(x)=x^2$ can be given codomain $\mathbb{R}$, or $[0,\infty )$, or $[-5,\infty )$. The rule alone does not fix it; the notation $f:A\to B$ does.

True — the range is $\{f(x):x\in A\}$, forced entirely by the domain and the rule. Unlike the codomain, we have no freedom to choose it.

False — $B$ (codomain) is only what we aim into; the range is what we hit. They coincide exactly when $f$ is onto.

True — every actual output is by definition one of the "declared possible" outputs, so the set of actual outputs sits inside the target set. It can never spill outside.

False — the range depends only on the domain and rule. Widening the codomain of $x^2$ from $[0,\infty )$ to $\mathbb{R}$ leaves the outputs untouched; it only changes whether the equation $\text{Range}=\text{Codomain}$ holds, i.e. whether the function is still onto.

True — fewer allowed inputs means fewer possible outputs, so the range can only shrink or stay the same, never grow. Removing $x=5$ can never add an output.

False as stated — that is a condition, not a set. The domain is the set $\mathbb{R}\setminus \{3\}=(-\infty ,3)\cup (3,\infty )$; the condition merely describes it.

False — $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$ and $g:[0,5]\to \mathbb{R}$, $g(x)=x^2$ share a rule but differ in domain, so they are different functions with different ranges.

True — "range = codomain" is the definition of onto; there are no output slots left empty in $B$.

The inversion is correct — from $y=\sqrt{4-x^2}$ square both sides to get $y^2=4-x^2$, so $x^2=4-y^2$ and $x=\pm \sqrt{4-y^2}$ — but the check is missing. The original rule only accepts $x\in [-2,2]$ (needed so $4-x^2\ge 0$), and $y=\sqrt{\cdot }$ is itself never negative, so $y$ is squeezed to $[0,2]$: the true range is $[0,2]$, not $[0,\infty )$. Skipping the domain check is forced because a value $y$ counts only if its preimage $x$ is actually an allowed input.

The solution $x=\pm \sqrt{y}$ only exists (as a real number) when $y\ge 0$; negative $y$ have no preimage, so the function misses all negatives and is not onto $\mathbb{R}$.

Being defined everywhere isn't enough — the quotient is undefined where the denominator is zero. Set $x-2\ne 0$, giving domain $\mathbb{R}\setminus \{2\}$.

Non-negativity is only half the story; the largest input $5$ caps the output at $25$. On this closed interval the range is $[0,25]$.

Inverting gives $x=\frac{2y+1}{y-1}$, which fails only when $y=1$ (division by zero, forcing a contradiction). So $y=1$ is unreachable and the range is $\mathbb{R}\setminus \{1\}$.

Sampling shows outputs that are in the range but never proves you found all of them or that gaps are truly missed. You need the existence argument: which $y$ make $f(x)=y$ solvable in the domain.

The symbol $\sqrt{x}$ denotes the non-negative root only — a function must give exactly one output per input. Allowing $\pm$ would violate the definition of a function itself.

Because "onto" (surjectivity) is precisely the statement range = codomain — without both concepts we couldn't even ask whether a function covers its target. In the mapping picture (Figure 1), onto means no output dot is left un-hit.

Solving $f(x)=y$ can produce an $x$ that is excluded from the domain (e.g. a forbidden value like $2$). Such a $y$ is not actually achieved, so skipping this check overstates the range.

A continuous curve cannot jump — to get from a low output to a high one it must pass through every height in between (the Intermediate Value idea sketched in Figure 3: the horizontal line at any in-between height must cut the graph). So it sweeps through all values from its lowest to its highest with no value skipped.

An inverse needs each output used exactly once (injectivity); restricting the domain (e.g. $x^2$ to $[0,\infty )$) removes the repeated outputs so the function becomes invertible. In a mapping diagram, injective means no two input arrows land on the same output dot.

The graph only lives above domain values and only reaches heights in the range, so knowing both boxes the curve into a rectangle before you plot a single point.

The graph shows only the actual outputs (the range); the codomain is a separate declaration about the target set, and you can't read a chosen target off the picture.

A single-element set $\{7\}$ — every input maps to $7$, so exactly one output is ever produced, no matter how large the domain.

Yes — the "empty function" with domain $\varnothing$ trivially satisfies "each input has one output" (there are no inputs to violate it), and its range is also empty.

Only if the domain is empty; otherwise at least one input produces at least one output. A non-empty domain forces a non-empty range.

No — $\frac{1}{x}=0$ has no solution for any real $x$, so $0$ is the one value the codomain $\mathbb{R}$ contains but the function never reaches.

$\{3\}$ — a one-element domain gives at most a one-element range, here exactly the single output $3$.

No — $f(-1)=f(1)=1$, so two distinct inputs share an output; it only becomes one-to-one once you restrict to $[0,2]$ (or $[-2,0]$).

Yes — the interval $[0,5]$ is closed, so $x=5$ is a legal input and $g(5)=25$ genuinely belongs to the range $[0,25]$.

Neither — the endpoints $0$ and $5$ are excluded as inputs, so the range becomes the open interval $(0,25)$; the extreme outputs are approached but never attained.