Date post: | 30-Dec-2015 |
Category: |
Documents |
Upload: | jean-waters |
View: | 226 times |
Download: | 2 times |
10.1
COMPOSITION OF FUNCTIONS
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Composition of Functions
• The function f(g(t)) is said to be a composition of f with g.
• The function f(g(t)) is defined by using the output of the function g as the input to f.
• The function f(g(t)) is only defined for values in the domain of g whose g(t) values are in the domain of f.
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Formulas for Composite FunctionsExample 1
Let p(x) = sin x + 1 and q(x) = x2 − 3. Find a formula in terms of x for w(x) = p(p(q(x))).
Solution
We work from inside the parentheses outward. First we find p(q(x)), and then input the result to p.
w(x) = p(p(q(x)))
= p(p(x2 − 3))
= p(sin(x2 − 3)) + 1
= sin(sin(x2 − 3) + 1) + 1.Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Composition of Functions Defined by Graphs
Example 3Let u and v be two functions defined by the graphs. Evaluate:
(a) v(u(−1)) (b) u(v(5)) (c) v(u(0)) + u(v(4))
Solution(a) To evaluate v(u(−1)), start with u(−1). From the figure, we see
that u(−1) = 1. Thus, v(u(−1)) = v(1). From the graph we see that v(1) = 2, so v(u(−1)) = 2.
(b) Since v(5) = −2, we have u(v(5)) = u(−2) = 0.(c) Since u(0) = 0, we have v(u(0)) = v(0) = 3. Since v(4) = −1, we
have u(v(4)) = u(−1) = 1. Thus v(u(0)) + u(v(4)) = 3 + 1 = 4.Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
3 2 1 1x
1
6
1 1 2 3 4 5 6x
3
1
1
3u(x) v(x)(-3,6)
(-1,1)
Composition of Functions Defined by TablesExample 2Complete the table. Assume that f(x) is invertible.SolutionWe will first look at g(f(2)). From the table we see that f(2) = 1. Therefore, we have g(f(2)) = g(1). Since g(1) = 1, we can fill in the entry for g(f(2)) the following way: g(f(2)) = g(1) = 1.To find g(2), we have to use information about g(f(2)). We first need to find a value of x such that f(x) = 2. That means that we are looking for f−1(2). From the table we see that f−1(2) = 0,or equivalently, f(0) = 2. Therefore, g(2) = g(f(0)). From the table we see that g(f(0) = 2. Thus, g(2) = 2.
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
x f(x) g(x) g(f(x))0 2 3 21 3 1 32 1
Decomposition of FunctionsExample 4Let h(x) = f(g(x)) = . Find possible formulas for f(x) and g(x).SolutionIn the formula h(x) = , the expression x2 + 1 is in the exponent. We can take the inside function to be g(x) = x2 + 1. This means that we can write
Then the outside function is f(x) = ex. We check that composing f and g gives h:
There are many possible solutions to Example 4. For example, we might choose f(x) = ex+1 and g(x) = x2. Then
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
12xe
12xe
)(12
)( xgx eexh
)()1())(( 12 2
xhexfxgf x
)())(( 11)( 2
xheexgf xxg
10.2
INVERTIBILITY AND PROPERTIES OF INVERSE
FUNCTIONS
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Definition of Inverse Function
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.
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Finding a Formula for an Inverse FunctionExample 3Find the inverse of the function f(x) = 3x/(2x + 1).SolutionFirst, we solve the equation y = f(x) for x:
y = 3x/(2x + 1) 2xy + y = 3x 2xy – 3x = – y x (2y – 3) = – y
x = – y/(2y – 3) x = y/(3 – 2y)
As before, we write x = f-1(y) = y/(3 – 2y).Since y is now the independent variable, by convention we rewrite the inverse function with x as the independent variable. We have y = f-1(x) = x/(3 – 2x).
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Noninvertible Functions: Horizontal Line Test
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.
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
The graph of q(x) = x2 fails the horizontal line test, so q(x) = x2 has no inverse.
3 3x
9
y q(x) = x2
Graphing A Function And Its InverseExample 5Let P(x) = 2x.
(a) Show that P is invertible. (b) Find a formula for P−1(x).(c) Sketch the graphs of P and P−1 on the same axes.(d) What are the domain and range of P and P−1?
Solution(a) Since P is an exponential function with base 2, it is always increasing, and
therefore passes the horizontal line test. (b) To find a formula for P−1(x), we solve for x in the equation 2x = y. We take the
log of both sides and simplify to get x = (log y)/(log 2) or switching variables: y = P-1(x) = (log x)/(log 2)
(c) Tables of values for P(x) and P-1(x). Interchanging the rows of P(x) gives P-1(x).
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
x -3 -2 -1 0 1 2 3P(x) 0.125 0.25 0.5 1 2 4 8
x 0.125 0.25 0.5 1 2 4 8P-1(x) -3 -2 -1 0 1 2 3
Graphing A Function And Its InverseExample 5 continuedLet P(x) = 2x.
(c) Sketch the graphs of P and P−1 on the same axes.(d) What are the domain and range of P and P−1?
Solution(b) y = P-1(x) = (log x)/(log 2) (c) From the tables of values for P(x) and P-1(x), we have the
graphs on the right.Notice how they are mirrorimages of each other throughthe line y = x
(d) The domain of P, an exponential function, is all real numbers, and its range is all positive numbers. The domain of P−1, a logarithmic function, is all positive numbers and its range is all real numbers.
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
3 1 4 8x
3
1
4
8y
y = x
P(x) = 2xP-1(x) = (log x)/(log 2)
The Graph, Domain, and Range of an Inverse Function
• Graph of f−1 is reflection of graph of f across the line y = x.
• Domain of f−1 = Range of f• Range of f−1 = Domain of f• Example:
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
x
y
f
f−1
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.
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Example 6(a)(a) Check that f(x) = x/(2x + 1) and f−1(x) = x/(1 − 2x) are inverse
functions of each other.Solution (a)
Similarly, you can check that f (f−1(x)) = x.
x
x
xx
xx
xx
xx
xx
xx
xf
xfxff
12112
122
1212
12
1221
12)(21
)())((1
A Property of Inverse Functions
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Example 6(b)f(x) = x/(2x + 1) and f−1(x) = x/(1 − 2x)(b) Graph f and f−1 on axes with the same scale. What are the
domains and ranges of f and f−1?Solution (b)
Properties of Inverse Functions
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
x
y
f−1(x) = x/(1 − 2x) with asymptotesy = -½ and x = ½
f(x) = x/(2x + 1)with asymptotesy = ½ and x = -½
The domain of f(x) is all real numbers except ½ and the range is all real numbers except - ½
The domain for f−1(x) is all real numbers except -½ and the range is all real numbers except ½
y =x
Restricting the Domain
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
A function that fails the horizontal line test is not invertible. For this reason, the function f(x) = x2 does not have an inverse function. However, by considering only part of the graph of f, we can eliminate the duplication of y-values. Suppose we consider the half of the parabola with x ≥ 0. This part of the graph does pass the horizontal line test because there is only one (positive) x-value for each y-value in the range of f.
x
y The graphs of f and f−1 are shown in the figure. Note that the domain of f is the the range of f−1, and the domain of f−1 (x ≥ 0) is the range of f.
f(x) = x2
xxf )(1
Inverse Trigonometric Functions
In Section 8.4 we restricted the domains of the sine, cosine, and tangent functions in order to define their inverse functions:
y = sin−1 x if and only if x = sin y and −π/2 ≤ y ≤ π/2
y = cos−1 x if and only if x = cos y and 0 ≤ y ≤ π
y = tan−1 x if and only if x = tan y and −π/2 < y < π/2 .
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
10.3
COMBINATIONS OF FUNCTIONS
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
The Difference of Two Functions Defined by Formulas: A Measure of Prosperity
Consider the population function P(t) = 2 (1.04)t and the number of people that a country can feed N(t) = 4 + 0.5 t where t is measured in years and both P and N represent millions of people.
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
50 100tyea rs
50
100
m i lli o n s o f p eop le
Shortages occur
(78.32,43.16)
47.226 80tyea rs2
14.865m i lli o n s o f p eop le
We explore this situation by plotting these two functions on the same graph and see when the population exceeds the number who can be fed.
We could also graph the difference N(t) – P(t) and observe when the maximum surplus occurs.
Point of maximum surplus
The Sum and Difference of Two Functions Defined by Graphs
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Example 1Let f(x) = x and g(x) = 1/x. By adding vertical distances on the graphs of f and g, sketch h(x) = f(x) + g(x) for x > 0.Solution The graphs of f and g are shown in the figure. For each value of x, we add the vertical distances that represent f(x) and g(x) to get a point on the graph of h(x). Compare the graph of h(x) to thevalues shown in the table.x ¼ ½ 1 2 4f(x) = x ¼ ½ 1 2 4g(x) = 1/x 4 2 1 ½ ¼ h(x) = f(x) + g(x) 4 ¼ 2 ½ 2 2 ½ 4 ¼
0 1 2 3 4 5 6x
1
2
3
4
5
6y
Note that as x increases, g(x) decreases toward zero, so the values of h(x) get closer to the values of f(x). On the other hand, as x approaches zero, h(x) gets closer to g(x).
y = h(x)
y = g(x)
y = f(x)
Example 6Find exactly all the zeros of the function
p(x) = 2x · 6x2 − 2x · x − 2x+1.SolutionWe rewrite the formula for p as
p(x) = 2x · 6x2 − 2x · x − 2x+1
= 2x(6x2 − x − 2) factoring out 2x
= 2x(2x + 1)(3x − 2) factoring quadratic
Since p is a product, it equals zero if one or more of its factors equals zero. But 2x is never equal to 0, so p(x) equals zero if and only if one of the linear factors is zero:
(2x + 1) = 0 or (3x − 2) = 0So x = −1/2 or x = 2/3
Factoring a Function’s Formula into a Product
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Tables for Example 3
The Quotient of Functions Defined by Tables: Per Capita Crime Rate
Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Year 2005 2006 2007 2008 2009 2010Years since 2005 0 1 2 3 4 5
Crimes in City A = NA(t) 793 795 807 818 825 831
Crimes in City B = NB(t) 448 500 525 566 593 652
Population of City A = PA(t) 61,000 62,100 63,220 64,350 65,510 66,690
Population of City B = PB(t) 28,000 28,588 29,188 29,801 30,427 31,066
rA(t) = NA(t) / PA(t) 0.013 0.0128 0.01276 0.01271 0.01259 0.01246rB(t) = NB(t) / PB(t) 0.016 0.01749 0.01799 0.01899 0.01949 0.02099
We see that between 2005 and 2010, City A has a lower per capita crime rate than City B. Furthermore, the crime rate of City A is decreasing, whereas the crime rate of City B is increasing. Thus, even though the table indicates that there are more crimes committed in City A, it appears that City B is, in some sense, more dangerous. The table also tells us that, even though the number of crimes is rising in both cities, City A is getting safer, while City B is getting more dangerous.