Transcript No Slide Title

Composition of Functions
• Suppose we have two money machines, both of which
increase any money inserted into them. Machine A doubles
our money while Machine B adds five dollars. The money
that comes out is described by a(x) = 2x for Machine A and
b(x) = x + 5 for Machine B, where x is the number of dollars
inserted. The machines can be hooked together so that the
money coming out of one machine goes into the other. There
are two ways of hooking up the machines which result in the
formulas shown below. The first formula is the composition
of b with a, while the second formula is the composition of
a with b.
$
A
B
$:
b(a(x))  a(x)  5  2x  5.
$
B
A
$:
a(b(x))  2b(x)  2(x  5)  2x  10.
• Which of these two compositions would you prefer? Why?
Decomposition of Functions
• Sometimes we reason backward to find the functions which
went into a composition. This process is called decomposition,
and there may be more than one way to do it.
• Problem. The vertex formula for the family of quadratic
functions is p(x) = a(x – h)2 + k. Decompose the formula into
three simple functions. That is, find formulas for u, v, and w
where p(x) = u(v(w(x))).
Solution. We work from the inner parentheses outward. Let
w(x) = x – h. Let v(x) = x2 and let u(x) = ax + k. Now, let’s
check our work: u(v(w(x))) = av(w(x)) + k = a(w(x))2 + k =
a(x – h)2 + k. Thus, we have correctly decomposed the formula
for p(x).
Definition of Inverse Function
• We have previously studied inverse functions. For
example, we defined the logarithm as the inverse function
of the exponential function. We now give a more careful
treatment of inverse functions.
• Suppose Q = f(t) is a function with the property that each
value of Q determines exactly one value of t. Then f has
an inverse function, f -1 and
f -1 (Q)  t if and only if Q  f(t).
If a function has an inverse, it is said to be invertible.
• Example. Solve sin x = 0.8 using an inverse function. A
calculator (set in radians) gives x = sin-1(0.8)  0.93.
Finding a formula for an inverse function
• In the graph below, we have P = f(t) = 20 + 0.4t, and we
-1
want to graphically find f (25).
Locate 25 on
the P-axis
P
Read off the value of t
corresponding to P = 25.
• If we continue with the example from the previous slide, we may
also solve algebraically for f -1 (25). The equation to be solved
is: 20 + 0.4t = 25, and we first subtract 20 from both sides to
obtain 0.4t = 5. Upon dividing by 0.4, we have t = 12.5 .
• This same algebraic procedure can be carried for a general P to
obtain the formula for f -1 . We must solve 20 + 0.4t = P for t.
We have 0.4t = P – 20, and thus,
P  20
t
 2.5P  50.
0.4
That is,
f -1 (P)  2.5P  50.
• Problem. Suppose we have a savings account which pays
4% interest compounded annually. The balance, in dollars,
in the account after t years is given by B = f(t) = 500(1.04)t.
The inverse function t = f -1(B) gives the number of years
for the balance to grow to $B. Find a formula for t = f -1(B).
Solution. We solve for t in terms of B in the given equation.
B  500(1.04)t
B
 (1.04) t
500
 B 
log
  t  log 1.04
 500 
log(B/500)
t  f (B) 
.
log 1.04
-1
The Horizontal Line Test
• If there is a horizontal line which intersects a function’s
graph in more than one point, then the function does not
have an inverse. If every horizontal line intersects a
function’s graph at most once, then the function has an
inverse. The graph of the function q(x) = x2, which is
shown below, fails the horizontal line test. Consider y = 4.
The line y = 4
intersects graph of
q(x) = x2 twice.
Evaluating an Inverse Function Graphically
Let u(x) = x3 + x + 1. The graph of u(x) is shown below, and
it looks as though u passes the horizontal line test. To find
u-1(4), we can proceed graphically or we can use Maple.
u-1(4)  1.213
evalf(solve(x^3+x+1=4,x));
1.213411663, -.6067058314 + 1.450612250 I, -.6067058314 - 1.450612250 I
The Graph, Domain, and Range of the Inverse Function.
• Suppose f is an invertible function. Then outputs of f are
inputs of f -1. Similarly, outputs from f -1 are inputs of f. It
follows that:
Domain of f -1 = Range of f
and
Range of f -1 = Domain of f.
• Example. The domain of f(x) = 10x is all real numbers
while its range is all positive real numbers. Of course, we
know that f -1(x) = log x, and the domain of f -1 is all
positive real numbers while its range is all real numbers.
• It is known that the graph of f -1 can be obtained by
reflecting the graph of f across the line y = x.
A Property of Inverse Functions
• If y = f(x) is an invertible function and y = f -1(x) is its inverse,
then
• f -1(f(x)) = x for all values of x for which f(x) is defined,
• f(f -1(x)) = x for all values of x for which f -1(x) is defined.
• Example. f(x) = x/(1+x) and f -1(x) = x/(1–x). We verify that
the first property listed above holds for these functions:

f(x)
x/(1  x)
f (f(x)) 


1  f(x) 1  x/(1  x)
-1
x/(1  x)

x/(1  x)

 x.
(1  x  x)/(1  x) 1 /(1  x)
Restricting the domain
• A function which fails the horizontal line test is not
invertible. For this reason, f(x) = x2 does not have an
inverse function. However, if we restrict the domain of f
to the set of nonnegative x values, then the restricted graph
does pass the horizontal line test. Thus, f(x) = x2 does
have an inverse on its restricted domain, x  0.
• If we solve y = x2 for x, where x  0, we have that x = y .
We note that y is defined to be the nonnegative value
whose square is y. Thus, f(x) = x2, restricted to x  0 has
the inverse function f -1(x) = x . We note that the domain
of f -1 is also x  0. Do you see why?
The graph of f(x) = x2 restricted to x  0, and the graph of
f -1(x) = x .
Combinations of Functions
• Like numbers, functions can be combined using addition,
subtraction, multiplication, and division. For addition, the sum
of two functions is defined by adding the values of the two
functions at each value of x where both functions are defined.
Subtraction, multiplication, and division of two functions are
handled similarly.
• Example. Let f(x) = x and g(x) = 1/x. Define h(x) = f(x) + g(x),
for x > 0. Complete the table below for the values of h(x).
x
0.25
0.50
1.00
2.00
4.00
f(x) = x
0.25
0.50
1.00
2.00
4.00
g(x) = 1/x
4.00
2.00
1.00
0.50
0.25
?
2.50
?
?
?
h(x) = f(x)+g(x)
Graph of h(x) = f(x) + g(x), where f(x) = x and g(x) = 1/x, x>0.
y
x
Factoring a Function’s Formula into a Product
• It is often useful to express a given function as a product of
functions.
• Example. Find exactly all the zeros of
p(x) 2x  6x 2  2x  x  2x 1.
• We can factor out a common factor of 2x:
p(x) 2x (6x 2  x  2)
• Next, we factor the quadratic:
p(x)  2x (2x  1)(3x  2)
• The zeros are
1
2
x ,x
2
3
since 2x is never zero.
A function defined as a quotient of two functions
• If we let f(x) = sin x and g(x) = cos x, then
f(x) sin x
tan x 

.
g(x) cos x
• The quotient function has all real numbers except odd
multiples of  /2 as its domain. That is, the values where
g(x) = cos x is zero must be excluded from the domain.
y = tan x
The sum of two even functions is even
Example.
f(x) 2cos(x),g(x)  12 x 2 .
f(x) g(x)
The product of two odd functions is even
• Example.
f(x)  x, g(x)  sin(x).
f(x) g(x)
Summary of Compositions,Inverses, & Combinations of Functions
• Two functions form a new function when the output of one
becomes the input of the other--this is composition.
• Decomposition of functions was also introduced.
• Suppose Q = f(t) is a function with the property that each value
of Q determines exactly one value of t. Then f has an inverse
function,
-1
f (Q)  t if and only if Q  f(t).
• We discussed: Finding a formula for f -1, the horizontal line test,
domain and range of f -1, and the relation of the graphs of f and f -1.
• The compositions of f and f -1 were formed in both possible ways.
• Restricting the domain of f so that the resulting function has an
inverse was investigated.
• Combining functions using addition, subtraction, multiplication,
and division was studied.