10.1 COMPOSITION OF FUNCTIONS

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1 0. COMPOSITION OF FUNCTIONS

2 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.

3 Eample Formulas for Composite Functions Let p() = sin + and q() = 3. Find a formula in terms of for w() = p(p(q())). We work from inside the parentheses outward. First we find p(q()), and then input the result to p. w() = p(p(q())) = p(p( 3)) = p(sin( 3)) + = sin(sin( 3) + ) +.

4 Composition of Functions Defined by Tables Eample Complete the table. Assume that f() is invertible. f() g() g(f()) We will first look at g(f()). From the table we see that f() =. Therefore, we have g(f()) = g(). Since g() =, we can fill in the entry for g(f()) the following way: g(f()) = g() =. To find g(), we have to use information about g(f()). We first need to find a value of such that f() =. That means that we are looking for f (). From the table we see that f () = 0, or equivalently, f(0) =. Therefore, g() = g(f(0)). From the table we see that g(f(0) =. Thus, g() =.

5 Composition of Functions Defined by Graphs Functions Modeling Change: Eample 3 Let u and v be two functions defined by the graphs. Evaluate: (a) v(u( )) (-3,6) (-,) 3 6 u() (a) To evaluate v(u( )), start with u( ). From the figure, we see that u( ) =. Thus, v(u( )) = v(). From the graph we see that v() =, so v(u( )) =. 3 v()

6 Composition of Functions Defined by Graphs Functions Modeling Change: Eample 3 Let u and v be two functions defined by the graphs. Evaluate: (b) u(v(5)) (-3,6) (-,) 3 6 u() v() (b) Since v(5) =, we have u(v(5)) = u( ) = 0.

7 Composition of Functions Defined by Graphs Functions Modeling Change: Eample 3 Let u and v be two functions defined by the graphs. Evaluate: (c) v(u(0)) + u(v(4)) (-3,6) (-,) 3 6 u() v() (c) Since u(0) = 0, we have v(u(0)) = v(0) = 3. Since v(4) =, we have u(v(4)) = u( ) =. Thus v(u(0)) + u(v(4)) = 3 + = 4.

8 Decomposition of Functions Eample 4 Let h() = f(g()) = e. Find possible formulas for f() and g(). In the formula h() = e, the epression + is in the eponent. We can take the inside function to be g() = +. This means that we can write h( ) f ( g( )) e f ( e ) g ( ) Then the outside function is f() = e. We check that composing f and g gives h: e h( ) There are many possible solutions to Eample 4. For eample, we might choose f() = e + and g() =. Then f ( g( )) e g ( ) e h( )

9 0. INVERTIBILITY AND PROPERTIES OF INVERSE FUNCTIONS

10 Definition of Inverse Function Suppose Q = f(t) is a function with the property that each value of Q determines eactly one value of t. Then f has an inverse function, f, and f (Q) = t if and only if Q = f(t). If a function has an inverse, it is said to be invertible.

11 Finding a Formula for an Inverse Function Eample 3 Find the inverse of the function f() = 3/( + ). First, we solve the equation y = f() for : y = 3/( + ) y + y = 3 y 3 = y (y 3) = y = y/(y 3) = y/(3 y) As before, we write = f - (y) = y/(3 y).

12 Noninvertible Functions: The Horizontal Line Test 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. y The graph of q() = fails the horizontal line test, so q() = has no inverse q() =

13 Graphing A Function And Its Inverse Eample 5 (a) Let P() =. (a) Show that P is invertible. (a) Since P is an eponential function with base, it is always increasing, and therefore passes the horizontal line test. 8 y y = P() = 4 P - () = (log )/(log )

14 Graphing A Function And Its Inverse Eample 5 (b) Let P() =. (b) Find a formula for P (). (b) To find a formula for P (), we solve for in the equation = y. We take the log of both sides and simplify to get = (log y)/(log ) or switching variables: y = P - () = (log )/(log )

15 Eample 5 (c) Let P() =. Graphing A Function And Its Inverse (c) Sketch the graphs of P and P on the same aes. (c) From the tables of values for P() and P - (), we have the graphs on the right. Notice how they are mirror images of each other through the line y = P() P - () y y = 4 P() = P - () = (log )/(log )

16 Graphing A Function And Its Inverse Eample 5 (d) Let P() =. (d) What are the domain and range of P and P? 8 y y = (d) The domain of P, an Eponential function, is all real numbers, and its range is all positive numbers. 4 P() = P - () = (log )/(log ) The domain of P, a logarithmic function, is all positive numbers and its range is all real numbers. 3

17 Graphing A Function And Its Inverse Eample 5 Let P() =. (a) Show that P is invertible. (b) Find a formula for P (). (c) Sketch the graphs of P and P on the same aes. (d) What are the domain and range of P and P? (a) Since P is an eponential function with base, it is always increasing, and therefore passes the horizontal line test. (b) To find a formula for P (), we solve for in the equation = y. We take the log of both sides and simplify to get = (log y)/(log ) or switching variables: y = P - () = (log )/(log ) (c) Tables of values for P() and P - (). Interchanging the rows of P() gives P - () P() P - ()

18 Graphing A Function And Its Inverse Eample 5 continued Let P() =. (c) Sketch the graphs of P and P on the same aes. (d) What are the domain and range of P and P? (c) From the tables of values for P() and P - (), we have the graphs on the right. 4 Notice how they are mirror images of each other through the line y = (d) The domain of P, an eponential function, is all real numbers, and its range is all positive numbers. The domain of P, a logarithmic function, is all positive numbers and its range is all real numbers. 8 y y = P() = P - () = (log )/(log )

19 The Graph, Domain, and Range of an Inverse Function Graph of f is reflection of graph of f across the line y =. Domain of f = Range of f Range of f = Domain of f Eample: y y = f f

20 A Property of Inverse Functions If y = f() is an invertible function and y = f () is its inverse, then f ( f()) = for all values of for which f() is defined, f( f ()) = for all values of for which f () is defined.

21 Eample 6(a) (a) Check that f() = /( + ) and f () = /( ) are inverse functions of each other. f f f f ) ( ) ( )) ( ( A Property of Inverse Functions Functions Modeling Change: Similarly, you can check that f ( f ()) =. (a)

22 Eample 6(b) Properties of Inverse Functions f() = /( + ) and f () = /( ) (b) Graph f and f on aes with the same scale. What are the domains and ranges of f and f? (b) y f() = /( + ) with asymptotes y = ½ and = -½ y = f () = /( ) with asymptotes y = -½ and = ½ The domain of f() is all real numbers ecept ½ and the range is all real numbers ecept - ½ The domain for f () is all real numbers ecept -½ and the range is all real numbers ecept ½

23 Restricting the Domain A function that fails the horizontal line test is not invertible. For this reason, the function f() = 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 0. This part of the graph does pass the horizontal line test because there is only one (positive) -value for each y-value in the range of f. y f() = f ( ) The graphs of f and f are shown in the figure. Note that the domain of f is the range of f, and the domain of f ( 0) is the range of f. Functions Modeling Change:

24 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 if and only if = sin y and π/ y π/ y = cos if and only if = cos y and 0 y π y = tan if and only if = tan y and π/ < y < π/.

25 0.3 COMBINATIONS OF FUNCTIONS

26 The Difference of Two Functions Defined by Formulas: A Measure of Prosperity Consider the population function P(t) = (.04) t and the number of people that a country can feed N(t) = t where t is measured in years and both P and N represent millions of people. We eplore this situation by plotting these two functions on the same graph and see when the population eceeds the number who can be fed. millions of people Shortages occur (78.3,43.6) t years We could also graph the difference N(t) P(t) and observe when the maimum surplus occurs. millions of people Point of maimum surplus t years Functions Modeling Change:

27 The Sum and Difference of Two Functions Defined by Graphs Eample Let f() = and g() = /. By adding vertical distances on the graphs of f and g, sketch h() = f() + g() for > 0. The graphs of f and g are shown in the figure. For each value of, we add the vertical distances that represent f() and g() to get a point on the graph of h(). Compare the graph of h() to y the values shown in the table. 6 ¼ ½ 4 f() = ¼ ½ 4 g() = / 4 ½ ¼ h() = f() + g() 4 ¼ ½ ½ 4 ¼ Note that as increases, g() decreases toward zero, so the values of h() get closer to the values of f(). On the other hand, as approaches zero, h() gets closer to g() y = h() y = g() y = f() Functions Modeling Change:

28 Factoring a Function s Formula into a Product Eample 6 Find eactly all the zeros of the function p() = 6 +. We rewrite the formula for p as p() = 6 + = (6 ) factoring out = ( + )(3 ) factoring quadratic Since p is a product, it equals zero if one or more of its factors equals zero. But is never equal to 0, so p() equals zero if and only if one of the linear factors is zero: ( + ) = 0 or (3 ) = 0 So = / or = /3

29 The Quotient of Functions Defined by Tables: Per Capita Crime Rate Tables for Eample 3 Year Years since Crimes in City A = N A (t) Crimes in City B = N B (t) Population of City A = P A (t) 6,000 6,00 63,0 64,350 65,50 66,690 Population of City B = P B (t) 8,000 8,588 9,88 9,80 30,47 3,066 r A (t) = N A (t) / P A (t) r B (t) = N B (t) / P B (t) We see that between 005 and 00, 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. Functions Modeling Change: