Reference Sheet: Programme F.10 — Functions

Rules, procedures and drill, in the order of the book's programme. Every rule links to the unit in the lesson that derives it — if a rule here looks arbitrary, the link is where it stops being arbitrary.

Notation, as in the book: domains and ranges are written as inequalities (-2\le x<3, -\infty<x<\infty), and a composition is written b[a(x)] with square brackets for the inner call.

1. Input, rule and output

From Unit 01.

A function is a rule that processes an input number x into an output number y=f(x). The box diagram shows it as a box with the instruction inside:

"y is a function of x" is the usual reading, but strictly the function is the rule f (like a calculator key); y is the number it produces.

Examples. y=\frac1x: box ^(-1), the reciprocal. y=x-6: box -6. y=4x: box x4. y=x^{1/3}: box ^1/3, the cube root. At x=8 these give 0.125, 2, 32 and 2.

2. Functions are rules, but not all rules are functions

From Unit 01.

A rule is a function when every input produces a unique output: it is single valued. Many inputs may share one output; one input may not have two outputs.

Vertical-line test. A graph is the graph of a function exactly when no vertical line meets it more than once. One vertical line meeting it twice is enough to disqualify it.

Even roots: two conventions.

When checking against the book's answers, read an even root written as a fractional power as \pm.

Examples. y=2x^3-x: a function. x^2+y^2=16 gives y=\pm\sqrt{16-x^2}; at x=0, y=\pm4: not a function. y=x^{2/3}=\left(x^{1/3}\right)^2: odd root, a function in both conventions. y=x^{3/4}: a function on x\ge0 here, "not a function" in the book.

3. Domain and range

From Unit 02.

• Domain: all the inputs the function may process. It may be stated (-5\le x<4); if none is stated it is all finite x, -\infty<x<\infty, less any point where the rule is undefined.
• Range: all the outputs actually produced from the domain.
• Finding the range: sweep the domain. Check the end values, every turning point inside the domain, and every asymptote. An excluded end value (<) gives an excluded output.
• A function that only rises (or only falls) on the domain takes its range from its end values.

Range and co-domain. Here the co-domain is the set the outputs are declared to lie in, and the range is the part actually produced. Stroud writes "range (or co-domain)" and treats the words as synonyms: read the book's co-domain as range.

Examples. y=x^3 on -2\le x<3: rising, so -8\le y<27. y=x^2 on -1\le x<3: the ends give 1 and 9, but the turning point x=0 gives 0, so 0\le y<9. y=2x+1 on -2<x\le4: -3<y\le9. y=x^4, no domain stated: -\infty<x<\infty, 0\le y<\infty. y=\sqrt{1-x^2}: domain -1\le x\le1, range 0\le y\le1.

4. Asymptotes and split ranges

From Unit 02.

A point where a denominator is zero is removed from the domain. At a vertical asymptote the output runs off to \pm\infty, and the range may split into two pieces. Sweep each piece of the domain separately.

Example. y=\frac1{x-3} on 0\le x\le5. Domain: 0\le x<3 and 3<x\le5. On the left piece x-3 runs from -3 up to 0, so y runs from -\frac13 down to -\infty. On the right piece x-3 runs from 0 up to 2, so y falls from +\infty to \frac12. Range: -\infty<y\le-\frac13 and \frac12\le y<\infty.

5. Functions and the arithmetic operations

From Unit 02.

The combination is defined on the common domain, where both f and g are defined. A quotient also excludes every x where its denominator is zero.

Example. f(x)=3x on -4\le x\le2 and g(x)=x+1 on -2<x\le5. Common domain: -2<x\le2. Then f+g=4x+1 on -2<x\le2, with range -7<y\le9. And \frac fg=\frac{3x}{x+1} on -2<x\le2, x\ne-1.

6. Inverses of functions

From Unit 03.

Reverse the flow through the box diagram: the output becomes the input, and the original input is retrieved as the output. The reversed rule is the inverse of the function, written f^{-1} or \operatorname{arc}f.

f^{-1} is not the reciprocal: f^{-1}(x)\ne\frac1{f(x)}.

Examples. f(x)=x-9 has f^{-1}(x)=x+9: f(10)=1 and f^{-1}(1)=10, whereas \frac1{f(1)}=\frac1{-8}=-0.125. f(x)=\frac x7 has f^{-1}(x)=7x.

7. Inverse operations

From Unit 03.

Swap and solve (an alternative, not in the book) gives the same inverse: write y=f(x), swap x and y, solve for y.

Examples. 6x\to\frac x6; x^3\to x^{1/3}; \frac x2\to2x. By swap and solve, y=x^5 becomes x=y^5, so y=x^{1/5}.

8. Self-inverse functions

From Unit 03.

A function equal to its own inverse: f(x)=x, and f(x)=\frac1x, since \frac1{1/x}=x.

Example. \frac1x sends 4\to0.25 and 0.25\to4.

9. Graphs of inverses

From Unit 04.

Swapping the coordinates of every point, (a,b)\to(b,a), reflects the graph in the line y=x: the segment from (a,b) to (b,a) is perpendicular to y=x and its midpoint \left(\frac{a+b}2,\frac{a+b}2\right) lies on it. The graph of f^{-1} is the graph of f reflected in y=x; the two are "reflection symmetric about y=x".

Example. (3,0.5) on the graph of f puts (0.5,3) on the graph of f^{-1}. Points on y=x itself are fixed.

10. The graphs of y = x^3 and its inverse

From Unit 04.

Plot y=x^3 from a table; swap the two columns (on a spreadsheet: copy, then paste the values into swapped columns) and plot again to get y=x^{1/3}. Add the line y=x: the curves are mirror images and cross at (-1,-1), (0,0) and (1,1).

Every horizontal line meets y=x^3 once, so its inverse x^{1/3} is a function for every real x.

11. The inverse of a function and the inverse function

From Unit 04.

• The inverse of the function: the whole reflected curve, which may be many-valued.
• The inverse function: what is left after removing branches of the inverse until it is single valued.

Reflecting y=x^2 gives the sideways parabola y=\pm\sqrt x: not a function (x=4 gives \pm2). Removing the lower branch leaves y=\sqrt x, the inverse function. Removing the upper branch instead leaves y=-\sqrt x, the inverse function of x^2 on x\le0.

Horizontal-line test (beyond the book): f has an inverse function on a domain exactly when no horizontal line meets its graph more than once there.

Example. The inverse of y=x^6 is y=\pm x^{1/6}. The book answers "x^{1/6}, not a function" (even root, two-valued); in these notes x^{1/6} is the non-negative root, the inverse function of x^6 on x\ge0.

12. The graph of y = x^4 − x^2 + 1 and its inverse

From Unit 04.

The curve is a W: a local peak 1 at x=0, dips to \frac34 at x=\pm0.7071. The line y=0.9 meets it four times (x=\pm0.3357, \pm0.9420), so its reflection is many-valued: the inverse of the function is not the inverse function. Keeping a piece on which the curve only rises, such as x\ge0.7071, gives one.

13. Function of a function

From Unit 05.

Chain the boxes: the output of a is the input of b.

The algebraic order is the reverse of the diagram order: a acts first but is written nearest to x, on the right.

Examples. A calculator: 4, reciprocal, square: 4\to0.25\to0.0625, so f=b\circ a with a(x)=\frac1x, b(x)=x^2, f(x)=\left(\frac1x\right)^2. With a(x)=x-1 and b(x)=x^2: b[a(x)]=(x-1)^2, and b[a(4)]=9.

14. Order matters

From Unit 05.

b[a(x)] and a[b(x)] are in general different functions.

Example. a(x)=x^2, b(x)=x+1: b[a(x)]=x^2+1 but a[b(x)]=(x+1)^2. At x=3: 10 against 16.

15. Three-function compositions

From Unit 05.

Work from the innermost bracket outwards: c(b[a(x)]) means a, then b, then c.

Example. a(x)=x+1, b(x)=x^2, c(x)=2x. c(b[a(x)])=2(x+1)^2; a(b[c(x)])=(2x)^2+1=4x^2+1; b(c[a(x)])=\left(2(x+1)\right)^2=4(x+1)^2. At x=1: 8, 5 and 16.

16. Decomposition

From Unit 05.

To split f into its boxes, list the calculator key presses that evaluate it. A decomposition is not unique.

Examples. f(x)=(4x-1)^3: \times4, -1, cube, so f(x)=c(b[a(x)]) with a(x)=4x, b(x)=x-1, c(x)=x^3. (x-2)^6 is b[a(x)] with a(x)=x-2, b(x)=x^6, or c(b[a(x)]) with b(x)=x^2, c(x)=x^3.

17. Inverses of compositions

From Unit 06.

Reverse the flow through the whole chain: invert every box and reverse their order.

The inverse is a function only if every box has an inverse function.

Example. f(x)=2x^3+7: boxes cube, \times2, +7. Reversed: -7, \div2, cube root: f^{-1}(x)=\left(\frac{x-7}2\right)^{1/3}. Check: f(2)=23 and f^{-1}(23)=8^{1/3}=2.

Traps

• Range from the end values alone. A ball with h=20t-5t^2 on 0\le t\le3 has end heights 0 and 15 m, but reaches 20 m at t=2. Check turning points inside the domain.
• Losing an open end. y=x^3 on -2\le x<3 has range -8\le y<27, not \le27: x=3 is never fed in.
• The union instead of the overlap. f+g exists only where both are defined; and a quotient also loses the zeros of its denominator.
• f^{-1} as a reciprocal. For f(x)=25.4x, f^{-1}(38.1)=1.5, not \frac1{25.4\times38.1}=0.001033.
• Reflecting in an axis. The inverse of y=x^3 passes through (8,2), not (2,-8): reflect in y=x.
• Calling a two-branched inverse a function. Reflecting x^2 gives \pm\sqrt x; prune a branch to get the inverse function.
• Reading b\circ a left to right. b\circ a means a first: with a(x)=x+3 and b(x)=4x, b[a(2)]=20, not a[b(2)]=11.
• Assuming composition commutes. VAT then a 5 voucher on 50 gives 55; voucher then VAT gives 54.
• Undoing a chain in the same order. F=1.8C+32 at 350 °F: 350\div1.8-32=162.4 (wrong); (350-32)\div1.8=176.67 °C (right).
• Mixing the root conventions. The book's "not a function" for y=x^{1/6} refers to the two-valued even root; with principal roots x^{1/6} is a function on x\ge0.

Self-check

1. Which of these are functions? (a) y=4-3x^2 (b) x^2+y^2=9, solved for y (c) y=x^{1/3}-x.
2. Find the range of y=x^2 on -3\le x<2.
3. Find the range of y=x^3-1 on -1<x\le2.
4. Find the domain and range of y=\frac1{x+1} on -3\le x\le1.
5. f(x)=x+1 on -1\le x\le3 and g(x)=x-2 on 0<x\le5. (a) Find h=f+g with its domain and range. (b) Find k=\frac fg with its domain.
6. By reversing the box, find the inverses of x-7, \frac x5 and x^5.
7. Show that f(x)=\frac1x is self-inverse. For f(x)=x+2, compare f^{-1}(5) with \frac1{f(5)}.
8. (3,27) lies on y=x^3. Which point lies on the graph of the inverse? Show that y=x is the perpendicular bisector of the segment joining them.
9. Is the inverse of y=x^2-4 a function? If not, prune it to give the inverse function.
10. a(x)=x-2, b(x)=3x, c(x)=x^2. Find b[a(x)] and a[b(x)] and evaluate both at x=4. Find c(b[a(x)]).
11. Decompose f(x)=2(x-1)^3+5 into four functions.
12. Invert f(x)=2(x-1)^3+5 by reversing the chain, and check at x=3.

Fully worked solutions

1. (a) Every x gives one value of 4-3x^2: a function. (b) y=\pm\sqrt{9-x^2}; at x=0, y=\pm3: two outputs, not a function. (c) The cube root is single valued for every real x: a function.

2. The ends give (-3)^2=9 (included) and 2^2=4 (excluded), but x=0 is in the domain and gives 0, the lowest value. Range: 0\le y\le9.

3. x^3-1 only rises, so the ends decide: (-1)^3-1=-2 (excluded) and 2^3-1=7 (included). Range: -2<y\le7.

4. x=-1 makes the denominator zero. Domain: -3\le x<-1 and -1<x\le1. On the left piece x+1 runs from -2 up to 0, so y runs from -\frac12 down to -\infty. On the right piece x+1 runs from 0 up to 2, so y falls from +\infty to \frac12. Range: -\infty<y\le-\frac12 and \frac12\le y<\infty.

5. Common domain: 0<x\le3. (a) h(x)=(x+1)+(x-2)=2x-1, which only rises: h(0)=-1 (excluded), h(3)=5. Range -1<h(x)\le5. (b) k(x)=\frac{x+1}{x-2} on 0<x\le3, and g(2)=0, so also x\ne2.

6. -7 reverses to +7: x+7. \div5 reverses to \times5: 5x. The fifth power reverses to the fifth root: x^{1/5}.

7. Reversing the box \frac1x gives \frac1x, because \frac1{1/x}=x; so f^{-1}=f. For f(x)=x+2: f^{-1}(x)=x-2, so f^{-1}(5)=3 (and indeed f(3)=5), while \frac1{f(5)}=\frac17.

8. (27,3). From (3,27) to (27,3) is 24 across and 24 down, the diagonal of a square, so it is perpendicular to y=x. The midpoint is \left(\frac{3+27}2,\frac{27+3}2\right)=(15,15), on y=x.

9. Swap and solve: x=y^2-4, so y=\pm\sqrt{x+4}. At x=5 that gives y=\pm3: two outputs, so the inverse is not a function. Keep the upper branch (equivalently, restrict y=x^2-4 to x\ge0): the inverse function is y=\sqrt{x+4}, x\ge-4.

10. b[a(x)]=3(x-2)=3x-6, and at x=4 it is 6. a[b(x)]=3x-2, and at x=4 it is 10. c(b[a(x)])=(3x-6)^2=9(x-2)^2.

11. Key presses: -1, cube, \times2, +5. So f(x)=d[c(b[a(x)])] with a(x)=x-1, b(x)=x^3, c(x)=2x, d(x)=x+5.

12. Reverse and invert: -5, \div2, cube root, +1: f^{-1}(x)=\left(\frac{x-5}2\right)^{1/3}+1. Check: f(3)=2\times2^3+5=21 and f^{-1}(21)=\left(\frac{16}2\right)^{1/3}+1=2+1=3.