Math 0861 Learning Centre
Function Notation
If you went to an accountant and asked how much you’d have to pay in taxes this year,
the accountant would probably respond by asking, “How much did you make last year?”
He can’t answer your question until he has some numbers from you to start his
calculation. Once he knows those numbers, he’ll do the same math that he’s done for
many other clients to figure out how much you owe in taxes.
Functions are like this. For any particular function, you give it a number, it performs
some calculations using that number, and it gives you a result. It doesn’t matter what
the number is at the start, the calculations will be the same.
One of the nice things about functions is we don’t have to know what the calculated
result is for us to talk about it. The tax accountant can talk to you about “the amount of
taxes you’ll have to pay on your income” even if he doesn’t know the dollar value yet.
He’s describing the answer by telling you which set of calculations he’s going to do (the
ones on the tax form) and the number he’s going to start with (your income).
In math, we like to label things with symbols and letters. In the same way that we can
represent a number by using a variable like x, we name functions with letters and
indicate what numbers they operate on. The symbol for a function usually looks like ƒ( ).
Inside the parentheses, we put the number or expression that we are going to apply the
function to. (This is called the argument of the function.) When we’re defining what
calculations the function performs, we use a variable like x. The function that adds 3 to
any number would look like this:
ƒ(x) = x + 3
We can then refer to the result when we apply this calculation to any number or
expression. If we want to write, “the answer we get when we add three to nine” we can
write ƒ(9). We would read these symbols as “f of nine”.
You might be wondering why we bother with this notation when we could easily write
“9 + 3”. As you progress through mathematics, you’ll encounter functions that are more
complicated than this one, where the easiest way to refer to a result is that it’s the
answer you get when you perform a particular calculations on a number. An example of
this that you’re already familiar with is fractions.
The answer to the problem 3 ÷ 7 is a repeating decimal. The length of the repeating part
is six digits. To write this number down exactly it would be tedious to have to memorize
those digits. We’d have to do that for all the other answers to division problems that
don’t work out evenly. Instead we refer to “the answer you get when you divide 3 by 7”
as a fraction: 3⁄7. Fraction form tells us what numbers we use in the calculation and what
operation to apply to them, even if it doesn’t convey its value. Fractions are a form of
function notation on two numbers. Change the numbers and you change the value.
© 2013 Vancouver Community College Learning Centre. AuthoredbybyEmily
Darren Rigby
Simpson
Student review only. May not be reproduced for classes.
Example 1: Given the function ƒ(x) = 5 − x, find (a) ƒ(3) (b) ƒ(1⁄2) (c) ƒ(8) (d) ƒ(q)
(e) ƒ(a + b).
Solution: To get the values, we can simply plug in and solve.
(a) ƒ(3) = 5 − [3] = 2.
(b) ƒ(1⁄2) = 5 − [1⁄2] = 41⁄2.
(c) ƒ(8) = 5 − [8] = −3.
(d) ƒ(q) = 5 − q.
(e) ƒ(a + b) = 5 − (a + b) = 5 − a − b.
That’s how it is with function notation. Whatever the argument of the function is, plug it
in. Some functions refer to their arguments more than once:
Example 2: Given the function ƒ(x) = |x + 3| + x², find (a) ƒ(6) (b) ƒ(−7) (c) ƒ(0.23)
(d) ƒ(z) (e) ƒ(x − 4).
Solution: We plug the argument of the function into the definition of the function
everywhere we see x:
(a) ƒ(6) = |[6] + 3| + [6]² = |9| + 36 = 9 + 36 = 45.
(b) ƒ(−7) = |[−7] + 3| + [−7]² = |−4| + 49 = 4 + 49 = 53.
(c) ƒ(0.23) = |[0.23] + 3| + [0.23]² = |3.23| + 0.0529 = 3.2829.
(d) ƒ(z) = |z + 3| + z²
(e) ƒ(x − 4) = |[x − 4] + 3| + [x − 4]² = |x − 1| + x² − 8x + 16.
In (e), we saw the variable from the function’s definition reused, but changed a bit. This
is nothing to worry about. We plug it in just like anything else.
Example 3: Given the function ƒ(x) = 4x + 7, solve (a) ƒ(x) = 11 (b) ƒ(x) = −5
(c) ƒ(x) = 0.
Solution: This is a harder problem, and it’s another reason to use function notation.
It lets us refer to an x value that will give the result that we want. To solve these
problems, we plug in a value for ƒ(x) in the definition equation and solve for x:
ƒ(x) = 4x + 7
(a) 11 = 4x + 7 → 11 − 7 = 4x + 7 − 7 → 4 = 4x → 4 ÷ 4 = 4x ÷ 4 → 1 = x
(b) −5 = 4x + 7 → −5 − 7 = 4x + 7 − 7 → −12 = 4x → −12 ÷ 4 = 4x ÷ 4 → −3 = x.
(c) 0 = 4x + 7 → 0 − 7 = 4x + 7 − 7 → −7 = 4x → −7 ÷ 4 = 4x ÷ 4 → −7⁄4 = x.
We can also use the output of one function as the input of another.
Example 4: Given the functions ƒ(x) = −x + 5 and g(x) = x² − 9, find (a) ƒ(g(2)) and
(b) g(ƒ(2)).
Solution: Order of operations says that we work from the inside out, so we find the
value of the function on the inside before we work on the one on the outside:
(a) g(2) = [2]² − 9 = 4 − 9 = −5; ƒ(g(2)) = ƒ( [−5] ) = −[−5] + 5 = 5 + 5 = 10.
(b) ƒ(2) = −[2] + 5 = 3; g(ƒ(2)) = g( [3] ) = [3]² − 9 = 9 − 9 = 0.
Notice that while we started with the same number, 2, in both cases, the answers
weren’t the same. The order in which the functions are applied makes a difference.
© 2013 Vancouver Community College Learning Centre.
Student review only. May not be reproduced for classes. 2
EXERCISES
A. Given the function ƒ(x) = x + 5, find:
1) ƒ(6) 5) ƒ(a)
2) ƒ(0) 6) ƒ(b − 3)
3) ƒ(25) 7) ƒ(y − 5)
4) ƒ(s) 8) ƒ(m² + n)
B. Given the function g(x) = 2x³ − 4x, find:
1) g(−2) 5) g(r)
2) g(0) 6) g(x + h)
3) g(10) 7) g(1⁄2k)
4) g(t) 8) g(7) − g(3)
C. Given the function ƒ(t) = |t − 4|, find:
1) ƒ(7) 5) ƒ(z)
2) ƒ(1) 6) ƒ(−t)
3) ƒ(0) 7) ƒ(Ay + B)
4) ƒ(x) 8) ƒ(j² + j)
D. Given the function ƒ(x) = 3x − 1, solve:
1) ƒ(x) = 8 5) ƒ(x) = 30
2) ƒ(x) = −4 6) ƒ(x) = 1⁄2
3) ƒ(x) = −13 7) ƒ(x) = 0.8
4) ƒ(x) = 0 8) ƒ(x) = 3p − 1
E. Given the functions g(x) = x² − 8 and h(x) = 4x + 1, find:
1) g(5) 5) g(g(3))
2) h(3) 6) h(h(1⁄5))
3) g(h(2)) 7) g(h(x))
4) h(g(−4)) 8) h(g(x))
F. Write a function in function notation that expresses the following instructions in
mathematical symbols:
1) Take a number, x, and multiply it by 5, then add 3 to the result.
2) Take a number, x, and add 3 to it, then multiply the result by 5.
3) Take a number, t, and square it, then add 4 times the number and 4 to the result.
4) Find the absolute value of 6 less than the square of a number, y.
5) Take a number, k, and subtract six from it, then cube the result.
© 2013 Vancouver Community College Learning Centre.
Student review only. May not be reproduced for classes. 3
G: In estimating how much lumber to buy (in units of board feet) for a basement
renovation, a contractor takes the number of feet of wall space in the basement, w,
multiplies by 2, and then adds 5 to account for waste.
1) Write the function, ƒ, the contractor is using to estimate the board feet, b, he
needs for a job. [Hint: if you have trouble writing the formula, do parts (2) and (3) of this
question and see what calculations you do to get the answer, then try part (1) again.]
Determine how many boards the contractor would order for a job with:
2) 60 ft. of wall space 4) 135 ft. of wall space
3) 28 ft. of wall space 5) 320 ft. of wall space
How much wall space would there be in a job if the contractor orders:
6) 105 board feet (bd ft) 8) 63 bd ft
7) 277 bd ft 9) 413 bd ft
10) What function could you write that solves the sort of problem in (G6) through
(G9)?
SOLUTIONS
A: (1) 11 (2) 5 (3) 30 (4) s + 5 (5) a + 5 (6) b + 2 (7) y (8) m² + n + 5
B: (1) −8 (2) 0 (3) 1960 (4) 2t³ − 4t (5) 2r³ − 4r
(6) 2x³ + 6x²h + 6xh² − 4x + 2h³ − 4h (7) 1⁄4k³ − 2k (8) 658 − 42 = 616
C: (1) 3 (2) 3 (3) 4 (4) |x − 4| (5) |z − 4| (6) |−t − 4| or |t + 4| (7) |Ay + B − 4|
(8) |j² + j − 4|
D: (1) x = 3 (2) x = −1 (3) x = −4 (4) x = 1⁄3 (5) x = 31⁄3 (6) x = 1⁄2 (7) x = 0.6
(8) x = p
E: (1) 17 (2) 13 (3) g(9) = 73 (4) h(8) = 33 (5) g(1) = −7 (6) h(9⁄5) = 41⁄5
(7) g(4x + 1) = 16x² + 8x − 7 (8) h(x² − 8) = 4x² − 31
F: (1) ƒ(x) = 5x + 3 (2) ƒ(x) = 5(x + 3) = 5x + 15 (3) ƒ(t) = t² + 4t + 4
(4) ƒ(y) = |y² − 6| (5) ƒ(k) = (k − 6)³ = k³ − 18k² + 108k − 216
[It would not be necessary to expand (k − 6)³ since it’s much easier to calculate in
its unexpanded form.]
G: (1) ƒ(w) = 2w + 5 (2) 125 boards (3) 61 boards (4) 275 boards (5) 645 boards
b 5
(6) 50 ft. (7) 136 ft. (8) 29 ft. (9) 204 ft. (10) ƒ(b) =
2
© 2013 Vancouver Community College Learning Centre.
Student review only. May not be reproduced for classes. 4
