How to read and transform an unfamiliar graph

A fictional process uses memory according to $R(t)=4-(t-2)^2$ on $0\le t\le4$ , with $t$ in hours and $R$ numerical volume in GB. What can we learn without calculating every minute? Read a graph through its domain, zeros, direction of change and bounds; transform it by changing coordinates.

Use functions, linear relationships and parabolas. This is invented, not measured memory use of real software.

Domain and axes first

The domain is $[0,4]$ , the zeros 0 and 4, and usage is positive between them. It increases from 0 to 2 hours and decreases from 2 to 4. “Increasing” compares any two inputs in that interval: if $t_1<t_2$ , then $R(t_1)<R(t_2)$ . It does not mean “positive”: a positive function can decrease.

When equality is allowed, use “nondecreasing,” $R(t_1)\le R(t_2)$ . A horizontal segment is then allowed; strict increase excludes it.

Here $0\le R\le4$ : the function is bounded below and above. Its maximum 4 occurs at 2; minimum 0 at the endpoints. A bound need not be attained: $1/x$ for $x>0$ stays above 0 and approaches it, but has no minimum of 0.

Hypothetical memory usage

Inverse function: swap coordinates

Top: R=4−(t−2)² on [0,4], maximum (2,4). Bottom: solid f(x)=2x+1, long dashes f⁻¹(x)=(x−1)/2, short dashes y=x. Points (1,3) and (3,1) are reflected. Lower axes have equal scales and abstract units; portions of lines are shown.

Symmetry needs the correct domain

An even function satisfies $f(-x)=f(x)$ ; an odd one satisfies $f(-x)=-f(x)$ , with a domain symmetric about zero. On the whole real line, $x^2$ is even and $x^3$ odd. The first graph reflects across the vertical axis; the second is unchanged by a half-turn about the origin.

Our $R(t)$ is symmetric about $t=2$ but is not even on $[0,4]$ : −1, for example, is outside its domain. Set $u=t-2$ and get $R=4-u^2$ on $[-2,2]$ , now an even relationship in $u$ . Changing the origin changes the parity conclusion.

What happens to coordinates

In $g(x)=f(x)+k$ , all points move up by $k$ when $k>0$ . In $g(x)=f(x-h)$ , they move right by $h$ when $h>0$ : the old input $u$ now occurs at $x=u+h$ . This explains the minus sign inside.

$g(x)=a f(x)$ multiplies vertical coordinates by $a$ ; $a<0$ also reflects across the horizontal axis. $g(x)=f(-x)$ reflects across the vertical axis. For $b>0$ , $g(x)=f(bx)$ divides horizontal coordinates by $b$ : $b=2$ compresses the graph horizontally by a factor of two. The domain changes under these same rules. Transformations, OpenStax §3.5.

Branch · Successive rulesOrder matters for both formula and domain.

Read composition from the inside

$f\circ g$ means $f(g(x))$ : first $g$ , then $f$ . For $f(x)=x+1$ and $g(x)=x^2$ , $f(g(x))=x^2+1$ , whereas $g(f(x))=(x+1)^2$ . At $x=2$ these give 5 and 9. The input must be in the domain of $g$ and the intermediate output $g(x)$ in the domain of $f$ . Composition, OpenStax §3.4.

Branch · The reverse pathAn inverse function is not the reciprocal 1/f.

When does the output uniquely recover the input?

An inverse $f^{-1}$ exists on the range of $f$ if every output there has exactly one original input. This is a one-to-one correspondence with its own range; graphically, every horizontal line intersects at most once. Then $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$ on their respective domains.

For numerical temperature $F(C)=9C/5+32$ , the reverse rule is $C(F)=5(F-32)/9$ : 20°C gives 68°F and returning gives 20°C. The formula gives no physical meaning to temperatures outside a physically admissible domain.

Swapping input and output takes $(x,y)$ to $(y,x)$ . On equal numerical scales, this reflects across $y=x$ . The lower panel shows $f(x)=2x+1$ and $f^{-1}(x)=(x-1)/2$ : $(1,3)$ becomes $(3,1)$ . The −1 notation does not mean $1/f(x)$ .

On the whole real line, $x^2$ has no inverse function: −2 and 2 both give 4. Restricting the input to $x\ge0$ gives inverse $\sqrt x$ on $x\ge0$ . Inverse functions, OpenStax §3.7.

Even an unfamiliar curve becomes readable when its domain, behavior and coordinate changes are separated. A reverse path requires one more question: does the output really identify a unique original input?