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 − g)(x) = f (x) − g(x)
(f + g)(x) = f (x) + g(x)
(f g)(x) = f (x)g(x)
f
f (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 g(x) 6= 0.
√
√
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) k/f
(c) f /h
(d) h − k
NOTE ABOUT (d): Even though h(x) − k(x) = 0, the domain is NOT all reals. In particular, h(x) and
k(x) both need to make sense!
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, there are TWO requirements.
1. First, the inner function f (x) has to make sense.
2. Next, the expression g(f (x)) has to make sense.
Thus, you figure out the domains of inner AND combined func, and take the values they have in common.
Ex 2: Using the same functions as in Example 1, find formulas for the following:
(a) f ◦ f
(b) f ◦ g
(c) g ◦ f
(d) g ◦ g
NOTE 1: f ◦ g 6= g ◦ f ... order matters!
NOTE 2: An interesting special case occurs 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. (There are
many possible answers... you’d have to have multiple choice to do this.)
x
Ex 4: Suppose f (x) = x−8
and g(x) = x7 . Find f ◦ g and 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!
Ex 5: Several values of two functions T and S are listed in the tables.
t
T (t)
8
6
7
1
6
7
1
8
4
3
x
S(x)
8
7
7
8
6
1
1
6
4
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
Remember: 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.) Analogously, there is a 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.
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, (x, y) is on f ’s graph iff (y, x) is on f −1 ’s graph.
To compute f −1 (a), think about working backwards: which value of x satisfies f (x) = a?
Ex: For instance, 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!”) For another, 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 on 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 .
(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. (Think: you had y in terms of x; now switch them!)
At the end, you’ll have x = f −1 (y). If necessary, you can switch the names of x and y at the end to write it as
y = f −1 (x).
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