1. 10.1 COMPOSITION OF FUNCTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

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. 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

3. Formulas for Composite Functions Example 1 Let p(x) = sin x + 1 and q(x) = x 2 − 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(x 2 − 3)) = p(sin(x 2 − 3)) + 1 = sin(sin(x 2 − 3) + 1) + 1. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

4. Composition of Functions Defined by Graphs Example 3 Let 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 u(x) v(x) (-3,6) (-1,1)

5. Composition of Functions Defined by Tables Example 2 Complete the table. Assume that f(x) is invertible. Solution We 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 xf(x)f(x)g(x)g(x)g(f(x)) 0232 1313 21

6. Decomposition of Functions Example 4 Let h(x) = f(g(x)) =. Find possible formulas for f(x) and g(x). Solution In the formula h(x) =, the expression x 2 + 1 is in the exponent. We can take the inside function to be g(x) = x 2 + 1. This means that we can write Then the outside function is f(x) = e x. We check that composing f and g gives h: There are many possible solutions to Example 4. For example, we might choose f(x) = e x+1 and g(x) = x 2. Then Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

7. 10.2 INVERTIBILITY AND PROPERTIES OF INVERSE FUNCTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

8. 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. 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

9. Finding a Formula for an Inverse Function Example 3 Find the inverse of the function f(x) = 3x/(2x + 1). Solution First, 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

10. 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) = x 2 fails the horizontal line test, so q(x) = x 2 has no inverse. q(x) = x 2

11. Graphing A Function And Its Inverse Example 5 Let P(x) = 2 x. (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 2 x = 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-3-20123 P(x)P(x)0.1250.250.51248 x0.1250.250.51248 P -1 (x)-3-20123

12. Graphing A Function And Its Inverse Example 5 continued Let P(x) = 2 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 (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 mirror images of each other through the 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 y = x P(x) = 2 x P -1 (x) = (log x)/(log 2)

13. 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 f f −1 y = x

14. 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

15. 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. A Property of Inverse Functions Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

16. 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 f −1 (x) = x/(1 − 2x) with asymptotes y = -½ and x = ½ f(x) = x/(2x + 1) with asymptotes y = 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

17. 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) = x 2 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. 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) = x 2

18. 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

19. 10.3 COMBINATIONS OF FUNCTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

20. 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 Shortages occur (78.32,43.16) 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

21. The Sum and Difference of Two Functions Defined by Graphs Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Example 1 Let 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 the values shown in the table. x¼½124 f(x) = x¼½124 g(x) = 1/x421½¼ h(x) = f(x) + g(x)4 ¼2 ½2 4 ¼ 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)

22. Example 6 Find exactly all the zeros of the function p(x) = 2 x · 6x 2 − 2 x · x − 2 x+1. Solution We rewrite the formula for p as p(x) = 2 x · 6x 2 − 2 x · x − 2 x+1 = 2 x (6x 2 − x − 2) factoring out 2 x = 2 x (2x + 1)(3x − 2) factoring quadratic Since p is a product, it equals zero if one or more of its factors equals zero. But 2 x 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) = 0 So 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

23. 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 200520062007200820092010 Years since 2005012345 Crimes in City A = N A (t)793795807818825831 Crimes in City B = N B (t)448500525566593652 Population of City A = P A (t)61,00062,10063,22064,35065,51066,690 Population of City B = P B (t)28,00028,58829,18829,80130,42731,066 r A (t) = N A (t) / P A (t)0.0130.01280.012760.012710.012590.01246 r B (t) = N B (t) / P B (t)0.0160.017490.017990.018990.019490.02099 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.