Algebra II Through Competitions Chapter 7 Function Composition and Operations

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1 . FUNCTIONS. Definition A function is a relationship between the independent variable x and dependent variable y. Each value of x corresponds exactly one value of y. Note two different values of x can have the same value of y, but one value of x cannot have two different values of y. For example, the price of fruits in a store is a function of fruit kind. One pound of apple (x ) and one pound of orange (x ) can have the same price $.99 per pound (y). But one pound of apple (x) cannot have two different prices at the same time (one price tag says $.99 per pound (y ) and price tag says $0.99 per pound (y ). x, x Example. (009 Indiana Algebra II) Let x, x. Then f(0) + f() + x, x f() is equal to A. 0 B. 9 C.0 D. E. None of these Solution: C. f(0) = 0. f() = + =. f() = + = 7. f(0) + f() + f() = = 0. Example. (009 Indiana Algebra II) If f(x ) = x f(x + ) +, what is the value of f()? 7 A. B. C. D. E. 7 7 Solution: B. f( ) = f( + ) + f() = f( + ) + f() = f() + f() = /7.

2 Example. (007 NC Algebra II) Let g(x) = x + b x + c and g() = 6. Determine g(5). A. 5 B. c 9 C..5c D..5c 0 E. c Solution: C. g() = 6 = () + b + c 0 c = b b = 5 c/. g(5) = (5) + b 5 + c = 5 + 5( 5 c/) + c = 5c/ + c =.5c. Example. (007 NC Algebra II ) If f () = and f (n + ) = (f (n)), what is the value of f ()? A.. B. 6. C. 6. D. 56. E. 65,56. Solution: D. f() = f( + ) = (f()) = and f() = f( + ) = ( f()) =6, and f() = f( + ) = (f()) = 6 = Domain and range A domain is all the possible values of x. A range is all the possible values of y. Example 5. (00 NC Algebra II) Determine the range of the function x. x Solution: E. Let y = F(x) = (x + )/x = + /x y Solving for x we get: x y. We see that y can never equal, but can equal anything else, the range of this function is {y Reals, y }... Vertical line test If each vertical line intersects a graph at no more than one point, the graph is the graph of a function. x

3 Figure (a) is the graph of y = x and it is not a function. Figure (b) is the graph of y = x and it is a function. (a) (b). SYMMETRY.. Properties Symmetric with respect to y-axis x-axis Origin y y y Even Function Nothing Odd function Example 6. (00 NC Algebra II) A function f is even if for each x in the domain of f, f (x) = f ( x). A function f is odd if for each x in the domain of f, f ( x) = f (x). Which of the following statement(s) is (are) true? I. The product of two odd functions is odd. II. The sum of two even functions is even. III. The product of an even function and an odd function is odd.

4 IV. If f is any function and the function F is defined by F(x) = [f (x) + f ( x)]/, then F is even. A. All statements are true. B. Only I, II and III are true. C. Only II, III and IV are true. D. Only II and III are true. E. Only III and IV are true. Solution: C. First note that I is false since the product of two odd functions, like f(x) = x and g(x) = x is cleary an even function. Next note that the next two are clearly true. The final option, that F( x) is always even bears checking. First, if f(x) is even, then F( x) =, which is even. If f(x) is odd, then F( x) = 0, which is a constant function, hence it is also even.. Graph of functions To graph Shift y f (x) by c y c Upward y c Downward y f ( x c) Left y f ( x c) Right Example 7. (005 NC Algebra II ) Suppose we are given the graph of y = f (x). Which functional expression below describes a graph that is reflected across the x-axis then shifted units to the right and 5 units up? A. f (x 5) B. f (x + ) 5 C. f ( x) + 5 D. f (x ) + 5 E. 5 f (x ).

5 Solution: E. When a graph is flipped over the x-axis, f (x) turns into f (x). To move a graph to the right, we must subtract from the x-value. To move a graph up we add to the final value, so the desired formula is f(x ) + 5 = 5 f(x ). Example 8. (00 NC Algebra II) If (, 5) is a point of the graph of y = f (x), then the graph of y = f (x ) contains the point (, c) where c equals A. B. C. D. 5 E. 6 Solution: D. If (, 5) is a point on the graph of y = f (x) then the point (, 5) will be the corresponding point on the graph of y = f(x ), since this graph is the result of sliding the original graph to the right units.. OPERATIONS ON FUNCTIONS Given two functions f(x) and g(x), we have the following operations Sum: (f + g) (x) = f(x) + g(x) Difference: (f g) (x) = f(x) g(x) Product: (f g) (x) = f(x) g(x) Quotient: (f/g) (x) = f(x)/g(x) g(x) 0. The domains of f + g, f g, f g) include all real numbers of the intersection of the domains of f and g. The domain of f/g includes all real numbers of the intersection of the domains of f and g for which g(x) 0.. COMPOSITE FUNCTIONS The composite function of f and g) is (f g)(x) = f(x) g(x) = f (g(x)). The domain of f g is the set of all numbers x in the domain g such that g(x) is in the domain of f. Example 9. (007 NC Algebra II) For f (x) = x and g(x) = x +, which has the greatest value? 5

6 A. f(g()). B. g(f ()). C. g(f ( )). D. f(g( 5)). E. g(0). Solution: A. g( x) f x f = x x x 9 and g( x ) ( x ) x g. So f(g()) = 80, g(f()) = 9, g(f()) =, f(g(5)) = 8, and g(0) =, so f(g()) has the largest value. Example 0. (999 NC Algebra II ) Let be an operation defined on functions such that: f g(x) f (g(x)) g(f (x)). If f(x) = x and g(x) = x +, find f g(x). A. x x B. x x + C. x + x D. x + x + E. none of the above. Solution: (D). f g(x) = (x + ) ((x ) + ) = x + x + (x + ) = x + x INVERSE FUNCTIONS 5. One-to-one function Horizontal line test: If each horizontal line intersects the graph of a function in no more than one point, the function is one-to-one. Figure (a) is the graph of y = r x and it is not one-to-one. Figure (b) is the graph of y = r x and it is one-to-one. Figure (a) Figure (b) 6

7 5. Inverse function f is a one-to-one function. g is the inverse function of f if (f g)(x) = x for every x in the domain of g and (g f)(x) = x for every x in the domain of f. g is written as f. Notes: () Do not confuse in f with a negative exponent. () A function f has the inverse function if and only if f is one-to-one. () The graph of f and f are symmetric with respect to the line y = x, that is, the graph of f is the image obtained by reflecting the graph of f about the line y = x. 5.. Steps in find an equation for f () Check if f defined by y = f(x) is a one-to-one function. () Solve for x. Let x = f (y) () Exchange x and y to get x = f (y). () Check that (f f )(x) = x and (f f)(x) = x. Note: If it is hard to explicitly solve for x, we just exchange x and y to get an expression that define the inverse function implicitly. Example. (000 NC Algebra II) Suppose f and g are both linear functions, with y = x + and f(g(x)) = x. Find the sum of the slope and the y-intercept of g. A. - B. C. 0 D. E. none of these Solution: C. If f(g(x)) = x, then f(x) and g(x) are inverse functions. To find the inverse of y = x +, switch x and y and solve for y. The result is that y = g(x) = (/)x +. The sum of the slope and y-intercept is / + / = 0 x Example. (00 NC Algebra II) Find, if possible, the inverse of. x 6x x x 6x A., x B., x C., x D., x x x x x E. no inverse exists 7

8 Solution: C. x To find the inverse, exchange x and y in the equation y and solve for y. Solving x x for y, xy x = y + xy y = x + y(x ) = x + y, x or x x, x. x Example. (005 NC Algebra II) If f ( ) x then x A. f(x) = x. B. f(x) = x 5. 5 C. x. D. E.. x 5 x. x Solution: E. If f x, then f (x ). Now let u = x, so x x and f ( u), or. u u 5 u 5 x 5 x u Example. (009 Indiana Algebra II) If f(x) is a linear function and the slop of y = f(x) is, what is the slope of y = f (x)? A. B. C. D. E. None of these Solution: D. Let y = f(x) = The slope is. x b f ( x) x b. y = x b x y b 8

9 PROBLEMS t Problem. (999 NC Algebra II) Given: f ( t), evaluate f. t c A. / ( + c) B. / ( + c) C. c D. c E. none of these Problem. (999 NC Algebra II ) Given f(x) = x + x 5, find t so that f (t) = 0. A..79 B..8 C. D..57 E. none of these Problem. (000 NC Algebra II) If A. x B. x C. x for all x > 0, then f(x) = x 8 D. E. x x h( x) Problem. (00 NC Algebra II ) Let h ( x ) for x,,, and, h ( ), find h(). A. B. 0 C. 5 D. E. 6 x Problem 5. (00 NC Algebra II) Given. Solve f. x x A. x = B. x = C. -/ D. x E. x Problem 6. (00 NC Algebra II ) If f (x) f (x ) x, and f (7), find f (5). A. 0 B. 8 C. 6 D. 8 E. Problem 7. (00 NC Algebra II) Given x and h( x) x, find (h f )( ). A. B. C. D. E. 8 8 Problem 8. (00 NC Algebra II) Let f be a linear function with the properties that f () f (), f () f () and f (5) = 5. Which of the following statements is true? A. f (0) < 0. B. f (0) = 0. C. f () < f (0) < f ( ) D. f (0) = 5 E. f (0) > 5 9

10 Problem 9. (005 NC Algebra II) Let f ( x ) = x x + 5, g ( x ) = x + a and f ( g ( x )) = x + c. Find the value of c. A. B. 7 C. a a D. 5 a E. 5. a Problem 0. (005 NC Algebra II) A function of the form has the x b following properties f() = and f (5) =. What is the value of f(0)? 5 5 A. B. C. D. E. None of these 5 7 Problem. (007 NC Algebra II) If f (x) =(x + 5) + 8, then what is the sum of the values of x for which f (x) =? A. -0. B. -7. C. 0. D. 0. E. 97. Problem. (007 NC Algebra II) Let f (x) be defined as the least integer greater than x/5. Let g(x) be defined as the greatest integer less than x/5. What is the value of g(8) + f (0)? A.. B.. C.. D.. E. 5. Problem. (007 NC Algebra II) A function f from the integers to the integers is defined as follows: n if n is odd f ( n). n/ if n is even Suppose k is odd and f (f (f (k))) = 7. What is the sum of the digits of k? A.. B. 6. C. 9. D.. E. 5. Problem. (007 NC Algebra II) If f( t ) = (5t + )/ t then f(t) =? 6t t t A. 6 t B. C. D. 5 E. t t t 5t Problem 5. (00 Indiana Algebra II) If f(x) = x and g(x) = 5 x, then f(g - (x)) is equal to: A. x + 0x + B. x 0x + C. 5 x D. 5 x E. none of these Problem 6. (009 Indiana Algebra II) If f(x) = x and f (x) is the inverse function of f(x), then f() f () is equal to A. B. C. D. 0 E. 0

11 Problem 7. (008 Indiana Algebra II) If f(x) = x + and g(x) = x + then g(f (x)) = A. x B. x 5 C. 9x + D. 9x + 7 E. 9x + 5 Problem 8. (00 UNC- Charlotte Algebra II) Suppose f(0) = and f(n) = f(n ) +. Let T = f(f(f(f(5)))). What is the sum of the digits of T? A. 6 B. 7 C. 8 D. 9 E. 0. Problem 9. (00 UNC- Charlotte Algebra II) Let the function f be defined by f(x) = x + 0. If m is a positive number such that f(m) = f(m) which of the following is true? A. 0 < m B. < m 8 C. 8 < m D. < m 6 E. 6 < m Problem 0. (0 Alabama Algebra II) Function f satisfies f(x) + f(5 x) = x for all real numbers x. The value of f() is A. 7/ B. /7 C. 5/ D. /5 E. None of these Problem. (00 Alabama Algebra II) Suppose that f(n + ) = f(n) + f(n ) for n =,,. Given that f(6) = and f() = 8, what is f() + f(5)? A. 8 B. 9 C. 0 D. E. Problem. (999 NC Algebra II) If f (x) x c, solve for x when (f f )(x) 0. A. x c B. x (c c ), x c c ) C. x c c D. x c c, x c c E. none of the above., x c c Problem. (008 Tennessee Algebra II) Which of the following functions is an odd function? A. x x 5 B. x C. f ( ) cos 5 D. x x x E. f ( x) x x Problem. Find the inverse of: x 8. x 8 A. x B. ( ) f x C. D. The inverse does not exist E. None of these. x 8 9x 8 Problem 5. (0 Illinois Algebra II) Let f ( x ). Then f (x) = kx w + f where k, w, and f are positive integers. Find the value of (k + w+ f ).

12 SOLUTIONS: Problem. Solution: E. t Let y. If the y's and the t's are switched, then solving for y arrives at the inverse. t y t y t f ( t) y f = c = c. t c c Problem. Solution: A. The input of f(x) will be the output of f (x) and the output of f(x) will become the input of f. This means that x = f (t) = 0. and t = f(x) = f(0.) =.79. Problem. Solution: E. The x in the denominator of f(x) = /(+x) represents half of the function s input. So f(x) = /(+/ x) = /(+x). f(x) = 8/(+x). Problem. Solution: D. h( x) h ( x ) h( x), so each term is just four-thirds greater than the one before it. So h ( ), h ( ) +, and h(). Problem 5. Solution: D. y To find the inverse, exchange the x and y and solve for y. Thus x xy x = y + y x xy y = x + y. From this we see that the function is its own inverse. If x we had graphed it, and noticed the symmetry about the line y = x, we could have drawn x the same conclusion. Thus f f x. Setting this equal to, x x x x x we have = x + = (x ) x =, so x. x

13 Problem 6. Solution: E. Since f(x) = f(x ) + x, f (7) = f (5) + 7 = f (5) + 7 f (5) =. Problem 7. Solution: C. To find f, x y x = y + y x f (x) = x / (h f )( ) = h f h h. 8. Then Problem 8. Solution: D. Since f is a linear function and f() f(), we know that the slope is greater than or equal to zero. Likewise, since f() f(), the slope is less than or equal to zero. Taken together, this means that the slope must be zero, making the function f(x) = 0x + 5, so f(0) = 5. Problem 9. Solution: A. f(g(x))= f(x + a)=(x + a) (x + a) + 5=x + (a )x+(a a + 5). Since we are told that f (g(x))= x +c, we know that a =0 and a a+ 5 = c, so a = 6 and a a+5= = = c. Problem 0. Solution: A. a a f ( ) = and f (5) f ( ) 5, so a = + b and a = 5 + 5b and b b 5 + b = 5 + 5b 8 = b b = and a = + () = 5, making f (0). 0 Problem. Solution: A. = (x + 5) + 8 = (x+5) ± = (x + 5), so x = or x = 7. So + ( 7) = 0. Problem. Solution: D. g(8) =, since is the least integer greater than 8/5. Similarly f (0) = 0, since 0 is the greatest integer less than 0/5. So g(8) + g(0) = + 0 =. Problem. Solution: B. Since k is odd, f(k) = k +. Since k + is even, f(k + ) = f(f(k)) = then 7 = f(f(f(k))) = k f = k k. If k is odd, +, which imlies that k = 5. This is not possible

14 k because f((f(5))) = f(f(8)) = f() =. Hence must be even, and k k 7 = f(f(f(k))) = f =, which implies that k = 05. Checking, we find that f((f(05))) = f(f(08)) = f(5) = 7. Hence the sum of the digits of k is = 6. Problem. Solution: A. Let x = t then t = x. f( t ) = 5 + t f(x) = 5 + ( x) = 6 x. f (t ) = 6 t. Problem 5. Solution: B. g - (x) = x + 5. We know that If f(x) = x, so f(g - (x)) = (g - (x)) = = ( x + 5) = x + 5 0x = x 0x +. Problem 6. Solution: E. Since f(x) = x, f() = ( ) =. Since f (x) = x, f ( ) = ( ) =. f() f () = =. Problem 7. Solution: A. Since f(x) = x +, f (x) = x. Since g(x) = x +, g(f (x)) = f (x) + = ( x ) + = x + = x. Problem 8. Solution: C. In fact, f(n) = n +, and T = f(f(f(f(5)))) = f(f(f())) = f(f(9)) = f(6) = 5. Problem 9. Solution: B. Note that f(m) = (m) + 0 = m + 0 and f(m) = (m + 0) = m It follows that m = 0, so < m 8. Problem 0. Solution: A. We know that f(x) + f(5 x) = x. f() + f(5 ) = () f() + f(5 ) = () () (): f() = 7 f() = 7/.

15 Problem. Solution: C. f(6) = f(5) + f() = f() + f() + f() + f() = 5f() + f() = () f() = f() + f() = f() + f() + f() = f() + f() = 8 () () (): f() =. Thus f() = 8 = 6. f() + f(5) = f() + f() + f() + f() = f() + f() + f() + f() + f() + f() = f() + f() = + ( 6) = 0. Problem. Solution: (C). f(f(x)) = f(x c ) = (x c ) c = 0 Let y = x and b = c. y yb ( b b) 0 b (b) ( b b) y = b b. x c c x c c Problem. Solution: D. For odd function, we have y. We test and we know that D is the answer. Problem. Solution: B. Let f(x) = y. x 8 y x 8 8 x y.. y 8 x y 8 x We switch the position of x and y: ( x 8 f ) x. Problem 5. Solution:. Let x + = X. 9x 8 9( X ) f ( x ) f ( X ). Let f(x) = y. 9( X ) f ( X ) 9( X ) y y 9( X ) X y. We switch the position of X and y: f ( X ) X. So k =, w = and f =. k + w+ f = =. 5

Section Properties of Rational Expressions

Section Properties of Rational Expressions 88 Section. - Properties of Rational Expressions Recall that a rational number is any number that can be written as the ratio of two integers where the integer in the denominator cannot be. Rational Numbers: