Real functions and their graphs

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Where does the graph of a real function sit, once every real input has been given one real output — and when does swapping the letters still leave one output for each new input?

The pairs sit on one line

Take the assignment $y = 2 x + 1$ . The extra assumption is that the pairs form a straight line. Slope and intercept already name that line on Lines in the plane. Here the line is only the picture of the pairs.

Drag x. The point stays on y=2x+1, the line through (0, 1) and (1, 3). The five fixed points are the rows x = 0, 1, 2, 3, 4.

The rows $x = 0, 1, 2, 3, 4$ give $y = 1, 3, 5, 7, 9$ . Those five points, and the point that moves when you drag, lie on the line through $(0, 1)$ and $(1, 3)$ . The graph is that set of points.

Call the assignment $f$ , and call the number it picks at $x$ the value $f (x)$ , so $y = f (x)$ . The domain $D (f)$ is the set of every $x$ that receives a value. The range $R (f)$ is the set of those values. The graph $G (f)$ is the set of pairs $(x, f (x))$ . A table and a graph fix the same assignment. On this line the domain and the range are every real number.

A larger input forces a larger value

On this line a larger input gives a larger value. Whenever $x < u$ ,

$f (x) < f (u) .$

This comparison is (0.27). The function is then strictly increasing.

Replace $f (x) < f (u)$ in turn by

$f (x) \leq f (u), f (x) > f (u), f (x) \geq f (u)$

and the function is nondecreasing, decreasing, or nonincreasing.

A flat step keeps the weak comparison

The next board freezes a flat step at height $1$ for $x < 1$ , then follows $y = x$ .

Drag either mark. Both stay at height 1 while the input stays below 1, so the strict rise fails on that step and the weak comparison still holds.

On that step two inputs share a value, so $x < u$ does not force $f (x) < f (u)$ . The weak comparison $f (x) \leq f (u)$ still holds, and the function is nondecreasing. Turn the strict comparison around and a falling graph is decreasing. Allow a flat step on the way down and it is nonincreasing.

Three further graphs are fixed by the picture alone.

The left half is the mirror of the right half. A positive height meets the curve twice.

An even graph is a mirror of itself across the vertical axis. The curve above is $y = x^{2}$ on the whole line, so one positive height meets it twice.

A half-turn about the origin lays the curve back on itself.

An odd graph falls back onto itself after a half-turn about the origin.

The same bend returns after a fixed step along the horizontal axis.

A periodic graph repeats the same bend after a fixed step along the horizontal axis.

Cut the domain, then swap the letters

The even graph is why a swap can fail. On the whole line, $y = 1$ meets $y = x^{2}$ at two places, so the swapped letters do not assign one $x$ to that $y$ .

Keep the extra cut $x \geq 0$ :

$y = x^{2}, x \geq 0.$

This is (0.28). For every $y \geq 0$ there is exactly one solution $x \geq 0$ . That solution is the nonnegative square root, written $x = y$ . Swapping the letters gives

$y = x .$

This is (0.29). Reflecting the graph of (0.28) across the diagonal produces the graph of (0.29).

Drag the input. It stays at least 0. The point on the square and the point on the square root sit opposite across the diagonal.

Any continuous increasing function inverts by the same reflection. Reading a power backwards is a logarithm. The same reflection produces $arcsin x$ and $arccos x$ .

The graph is the set of pairs. Increase and decrease are comparisons of those pairs. An inverse is the diagonal reflection after each value has been given a single preimage.