Exercises — Domain, codomain, range

Domain = the set we map from = $\{1,2,3\}$. Codomain = the set we map into, read off the arrow's right side = $\{a,b,c,d\}$. Range = the outputs that actually appear = $\{a,a,c\}$. A set never repeats elements, so this is $\{a,c\}$.

Notice $\{a,c\}\subseteq \{a,b,c,d\}$: the range sits inside the codomain, and $b,d$ are "declared but never produced". Not onto.

Legal inputs are exactly the elements of the domain ${\mathbb{R}}_{\ge 0}$.

• $-4$: negative, not in domain → illegal (you cannot take $\sqrt{-4}$ and stay real).
• $0$: $0\ge 0$ → legal, $\sqrt{0}=0$.
• $9$: $9\ge 0$ → legal, $\sqrt{9}=3$.

Legal inputs: $\{0,9\}$.

The only thing a fraction forbids is dividing by zero. So we need the denominator $\ne 0$: $x^2-9\ne 0⟹x^2\ne 9⟹x\ne 3\text{ and }x\ne -3.$ Why both signs? $x^2=9$ has two solutions, $x=3$ and $x=-3$ (a square erases the sign). Miss one and you leave a hidden hole. Domain $=\mathbb{R}\setminus \{-3,3\}=(-\infty ,-3)\cup (-3,3)\cup (3,\infty )$. Note we write a set, not the condition "$x\ne \pm 3$".

A real square root demands its inside be $\ge 0$: $5-x\ge 0⟹x\le 5.$ Domain $=(-\infty ,5]=\{x\in \mathbb{R}:x\le 5\}$. The endpoint $5$ is included (there $\sqrt{0}=0$, perfectly fine).

$g$ is a straight line with positive slope $2$, so it is increasing: bigger input gives bigger output, with no dips. On a closed interval an increasing continuous line sweeps every value between its endpoint outputs.

• Left endpoint: $g(-2)=2(-2)+1=-3$ (the minimum).
• Right endpoint: $g(3)=2(3)+1=7$ (the maximum). Range $=[-3,7]$.

A value $y$ is in the range iff $y=\frac{x}{x^2+1}$ has a real solution $x$. Clear the fraction (valid since $x^2+1$ is never $0$): $y(x^2+1)=x⟹y{x}^2-x+y=0.$ This is a quadratic in $x$ (with $y$ as a constant). A quadratic $a{x}^2+bx+c=0$ has a real root exactly when its discriminant $b^2-4ac\ge 0$ — that square-root part must not go imaginary. Here $a=y$, $b=-1$, $c=y$: $(-1{)}^2-4(y)(y)\ge 0⟹1-4y^2\ge 0⟹y^2\le \frac{1}{4}⟹-\frac{1}{2}\le y\le \frac{1}{2}.$ Edge case $y=0$: then the equation is linear, $-x=0$, giving $x=0$ — a valid input, so $y=0$ is in. Good, it lies inside our interval. Range $=\left[-\frac{1}{2},\frac{1}{2}\right]$. See the figure: the curve rises to a peak of $\frac{1}{2}$ and dips to $-\frac{1}{2}$, never escaping that band.

Set $y=\frac{2x-3}{x+1}$ and solve for $x$: $y(x+1)=2x-3⟹yx+y=2x-3⟹yx-2x=-3-y⟹x(y-2)=-(y+3).$

• If $y\ne 2$: $x=\frac{-(y+3)}{y-2}$, a genuine real number. We must still check this $x$ is a legal input, i.e. $x\ne -1$. Suppose $x=-1$: then $-1=\frac{-(y+3)}{y-2}$ gives $-(y-2)=-(y+3)$, i.e. $-y+2=-y-3$, i.e. $2=-3$ — false. So the forbidden input is never hit; every $y\ne 2$ is achievable.
• If $y=2$: the equation becomes $0=-(2+3)=-5$, impossible — no $x$ produces $y=2$. Range $=\mathbb{R}\setminus \{2\}$. The missed value $2$ is exactly the ratio of leading coefficients $\frac{2}{1}$ — the horizontal level the curve approaches but never touches.

Domain: need the inside $\ge 0$: $9-x^2\ge 0⟹x^2\le 9⟹-3\le x\le 3$. Domain $=[-3,3]$. Range: because of the leading $+\sqrt{}$, outputs are never negative, so $y\ge 0$. To find the largest $y$, note $y=\sqrt{9-x^2}$ is biggest when $9-x^2$ is biggest, which happens at $x=0$: $y=\sqrt{9}=3$. The smallest $y$ is $0$, at $x=\pm 3$. Every value between is swept continuously. Range $=[0,3]$. Graph: squaring $y=\sqrt{9-x^2}$ gives $x^2+y^2=9$ with $y\ge 0$ — the upper half of a circle of radius $3$. See the figure.

Complete the square — rewrite so $x$ appears once, revealing the lowest point: $x^2-4x+7=(x^2-4x+4)+3=(x-2{)}^2+3.$ A square $(x-2{)}^2\ge 0$, hitting $0$ at $x=2$ and growing without bound as $x$ moves away. So $h(x)\ge 3$, minimum $3$, no upper limit. Range $=[3,\infty )$. Codomain $=\mathbb{R}$. The gap is $(-\infty ,3)$ — every value below $3$ is declared but never produced. Not onto.

Try $f(x)=e^x+2$. The exponential $e^x$ (see Graphing Functions) is always strictly positive: $e^x>0$ for every real $x$, and it approaches $0$ as $x\to -\infty$ but never equals $0$. Therefore $e^x+2>2$ always, and as $x\to -\infty$ the output gets arbitrarily close to $2$ without touching it.

• Is $2$ excluded? We'd need $e^x=0$, impossible → $2\notin$ range. ✔ (open at $2$).
• Is every $y>2$ reached? Solve $e^x+2=y⟹e^x=y-2⟹x=\ln (y-2)$, defined whenever $y-2>0$, i.e. $y>2$. ✔ Range $=(2,\infty )$, exactly as required.

Composite formula: $(g\circ f)(x)=g(f(x))=\frac{1}{f(x)-1}=\frac{1}{(x^2+1)-1}=\frac{1}{x^2}.$ Domain: $g$ forbids its own input being $1$ (that makes its denominator $0$), i.e. we must avoid $f(x)=1$. But we also see the simplified form $\frac{1}{x^2}$ forbids $x=0$. Check both:

• $f(x)=1⟹x^2+1=1⟹x=0$. Same culprit. So the only excluded input is $x=0$. Domain $=\mathbb{R}\setminus \{0\}$. Range: for $x\ne 0$, $x^2>0$, so $\frac{1}{x^2}>0$, and it takes every positive value (as $x\to 0$ it blows up to $\infty$; as $x\to \pm \infty$ it shrinks to $0$ but never reaches $0$). Range $=(0,\infty )$.

On the domain $[0,\infty )$, $x^2$ produces exactly the outputs $[0,\infty )$: it is $0$ at $x=0$ and rises to every non-negative value. So the range is $[0,\infty )$. A function is onto iff codomain $=$ range. Hence choosing $B=[0,\infty )$ makes $f$ onto. With domain $[0,\infty )$ it is also one-to-one (each output from exactly one input, since we dropped the negative branch — see Injective Functions). Being one-to-one and onto, it is invertible, with $f^{-1}(y)=\sqrt{y},f^{-1}:[0,\infty )\to [0,\infty ).$

L1 — read the three sets straight off the arrow and the actual outputs. L2 — domain from "no ÷0, no √(negative)"; range of a monotone line from its endpoints. L3 — range of a rational: invert, use discriminant / solvability, watch the linear case. L4 — natural domain + geometric range (half-circle, parabola vertex via completing the square). L5 — build functions to hit a target range, chain composites, and tune the codomain for onto-ness.