Section 2.7: Function Arithmetic, Composition
Functions can be added, subtracted, etc., just like numbers! If f (x) and g(x) are functions, then we can
define four new functions as follows:
f
f (x)
(f + g)(x) = f (x) + g(x)
(f − g)(x) = f (x) − g(x)
(f g)(x) = f (x)g(x)
(x) =
g
g(x)
In other words, to add functions, just add their y-values, and so forth!
Domain of f + g, f − g, etc.: Both f (x) and g(x) need to make sense! Thus, you take the points that f
and g’s domains have in common. For f /g, you also make sure that the denominator is not zero.
Make sure to check the domains of f and g before you simplify their combination!
√
√
Ex 1: Suppose f (x) = x + 2, g(x) = x2 + 5, h(x) = x, and k(x) = x. Find formulas and domains for
each of the following.
(a) f + g
(b) f /g
(e) h − k
(f) h/k
(c) k/f
(d) f /h
NOTE for (e) and (f): Even though h(x) − k(x) = 0 and h(x)/k(x) = 1, the domain is NOT all reals; you
need to check the domains before simplifying.
Composition
This is a new, and very important, way to combine functions. The composite of g and f is
(g ◦ f )(x) = g(f (x))
Here, g is the outer function, and f is the inner function. You plug f (x) into g. Think “do f , then do g”.
Domain of g ◦ f : In order for (g ◦ f )(x) to make sense, you check the domains of two parts:
1. The inner function f (x)
2. The total expression g(f (x))
Note that you do not have to check the outer function on its own.
Ex 2: Using the same functions as in Example 1, find formulas for these. What kinds of functions are they?
(a) f ◦ f
(b) f ◦ g
(c) g ◦ f
(d) g ◦ g
NOTE 1: f ◦ g 6= g ◦ f ... order matters!
NOTE 2: Note what happens when one function is constant. For instance, if f (x) = 2 and g(x) = x + 2,
then f (g(x)) = f (x + 2) = 2 (basically, f “ignores” whatever you put in and outputs 2 regardless).
√
Ex 3: Suppose y = x + 5 + 2. Determine two functions f (x) and g(x) so that f (g(x)) = y.
(This would be a multiple-choice question, since there are many possible options.)
x
Ex 4: Suppose f (x) = x−8
and g(x) = x7 . Find (a) f ◦ g and (b) g ◦ f and their domains.
(First, note that f ’s domain is all x’s except 8, and g’s domain is all x’s except 0.)
Understanding Functions in Other Contexts
Functions aren’t always presented through formulas. We can read a function off of a graph or maybe from a
table of values. With a table, you look up the input in its row (or column) and find the corresponding output
in another row (or column).
Ex 5: Several values of two functions T and S are listed in the tables.
t
8 7 6 1 4
x
8 7 6 1 4
T (t) 6 1 7 8 3
S(x) 7 8 1 6 3
Find the expressions, if possible. (If it is not possible, enter NONE.)
(a) (S ◦ T )(7)
(b) (T ◦ T )(7)
(c) (S ◦ S)(7)
(d) (T ◦ S)(4)
Ex 6: Suppose g(t) is the area of a circle as a function of time, and f (A) is the radius of a circle as a
function of its area.
(a) Find a formula for f (A).
(b) Determine what the composite (f ◦ g)(t) represents.
Section 4.1: Inverse Functions
Recall a function’s graph passes the Vertical Line Test, meaning no vertical line strikes twice. (i.e. Each
x coordinate can only produce one y.) There is a similar Horizontal Line Test, saying that no horizontal line
strikes twice. (Hence, each y coordinate only comes from one x.)
A function passing the Horizontal Line Test is called one-to-one or invertible.
REMARK: Another way of saying y = f (x) is one-to-one is to say that when you solve for x, you don’t get
multiple solutions.
Ex 7: Is the function f (x) = 6x − 1 one-to-one? What about g(x) = (x + 3)2 − 4?
If f is invertible, then the inverse function f −1 is the function that swaps the roles of input and output.
Therefore,
y = f (x) ⇐⇒ x = f −1 (y)
In other words, (a, b) is on f ’s graph iff (b, a) is on f −1 ’s graph.
Computing f −1 (a): Think about working backwards... which value of x makes f (x) = a?
Sample: If f (1) = 3 and f is invertible, then f −1 (3) = 1. (i.e. “f sends 1 to 3, so f −1 sends 3 back to 1!”)
Sample 2: If A = f (x) computes the area of a square as a function of side length x, then x = f −1 (A)
computes the side length as a function of area.
Ex 8: Using the tables from Example 5, find the expressions, if possible.
(a) T −1 (1)
(b) S −1 (4)
(c) T −1 (S(7))
(d) S(S −1 (3))
NOTE: f −1 does not mean the same thing as f to the −1 power... it is not 1/f .
Important Properties of f −1
• Theorem of Inverse Functions: f and f −1 cancel each other when you compose either way.
f (f −1 (x)) = x and f −1 (f (x)) = x
• The graph of y = f −1 (x) is obtained by reflecting the graph of y = f (x) over the diagonal line y = x.
• When inverting, domain and range switch! Thus, the domain of f is the range of f −1 and vice versa.
Ex 9: The graph of a one-to-one function f is provided. (The labeled points are (0, 4) and (12, −4).)
(a) Sketch a graph of f −1 .
NOTE: If you draw both graphs and tilt your
paper (or your head) appropriately, you should
see that the two graphs are mirror images over
y = x.
(b) Find the domain and range of f .
(c) Find the domain and range of f −1 .
Recall our work with transforming a point on a graph. We can do the same with inverses! The key thing is:
if (a, b) is a point on y = f (x), then when using the inverse, start instead with (b, a). (Switch the coordinates!)
Ex 10: Suppose (a, b) is on the graph of y = f (x). Find the corresponding point on:
(a) y = f −1 (x)
(b) y = 2f −1 (x − 1) + 1
Computing f −1
To find the inverse of y = f (x), solve x in terms of y. At the end, you’ll have x = f −1 (y). If needed, you
then switch the variables names to get y = f −1 (x). (I do this LAST.)
Ex 11: For each function, compute the inverse function f −1 (x).
(a) f (x) = 2x + 1
(b) f (x) = x2 + 1 where x ≤ 0
(c) f (x) = 3x3 − 1
x+5
(d) f (x) = 2x−1