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1 Objective 2 Inverse Functions Illustrate the idea of inverse functions. f() = 2 + f() = Two one-to-one functions are inverses of each other if (f g)() = of g, and (g f)() = for all in the domain of f. for all in the domain We write f to denote the inverse function. Objective 2b How are the graphs of f and f related? If (a,b) is on the graph of = f(), then is on the graph of = f (). 77

2 Objective 2b Eample Select the graph of = f (). A function can be its own inverse. Consider Objective 2a Does ever function have an inverse? to be be the graph of a one- Graph of a function must pass the to-one function. 78

3 Which are one-to-one functions? {(,2),(,3),(5,4)} {(,2),(3,2),(4,5)} {(,2),(3,3),(4,5)} If a function is not one-to-one, restrict the domain in order to define an inverse function. (Recall intro to Obj 2.) Objective 2d Given a function, find the function rule for f. *************** Plan of Attack. Write for f() (to simplif the notation). 2. Solve for. For applied mathematicians, when units are usuall associated with the variables, ou have the inverse function. 3. For our College Algebra course, we will interchange and to write the inverse as a function of. 4. Write f () for (to return to function notation). *************** Find the function rule for f for each of the following. f() = 3 5 f() = (4 )

4 f() = f() = + 3 f() = Objective 22 Eponential Functions. f() = a, a > 0, a Does this define a function? Don t allow base to be negative because could be Don t allow base to be because for some. i.e., graph would be linear, not eponential. What s the domain? All reals? If so, we have to define what s meant b irrational eponents. For eample: 4 2 or 4 π We haven t worked with irrational eponents. Good News: The limiting processes of calculus guarantee that irrational eponents are defined, and line up as we want. That means, since 3 < π < 4, then a 3 < a π < a 4. That means, since < 2 < 2, then a < a 2 < a 2. 80

5 The eponential functions are classified into 2 groups, depending on the base. f() = a, a > f() = a, 0 < a < We will consider two specific cases to develop the concept. This is not an egrade eample; ou will not be making tables of values - ou will not be plotting points. ( Consider f() = 4 Consider f() = 4) for an eample of a > for an eample of 0 < a < Objective 22a Properties and Graphs of Eponential Functions f() = a, a > f() = a, 0 < a < 8

6 Objective 22b Graphing Eponential Functions with Reflections or Translations Don t. Don t. Use Obj 4! Select the graph that best represents the graph of each of the following. ( f() = 5 f() = 4) Which function best describes the graph shown? Which function best describes the graph shown? f() = (2.5) f() = (2.5) f() = (2.5) f() = (2.5) f() = (0.4) f() = (0.4) f() = (0.4) f() = (0.4) More Objective 22b Graphing Eponential Functions with Translations Don t. Don t. Use Obj 4! 82

7 Select the graph that best represents the graph of each of the following. f() = 4 3 f() = ( ) +2 5 Which function best describes the graph shown? Which function best describes the graph shown? f() = 6 +3 f() = 6 +3 f() = ( ) +2 5 f() = ( ) f() = (0.6) +3 f() = (0.6) +3 f() = ( ) 2 2 f() = ( )

8 Objective 22c The eponential function is f() = e because of so man areas of application. ( e e = lim n + ) n n Graph f() = e Evaluate e on a scientific calculator (the Mac calculator in lab class). Strontium 90 is a radioactive material that decas over time. The amount, A, in grams of Strontium 90 remaining in a certain sample can be approimated with the function A(t) = 225e 0.037t, where t is the number of ears from now. How man grams of Strontium 90 will be remaining in this sample after 7 ears? $8,000 is invested in a bond trust that earns 5.9% interest compounded continuousl. The account balance t ears later can be found with the function A = 8000e 0.059t. How much mone will be in the account after 6 ears? 84

9 Objective 22d Solving eponential equations when we can obtain the same base. Eponential functions are one-to-one; that means: if and onl if Rewrite each side (if needed) in terms of a common base; use the smallest base possible. Be sure to replace equals. Solve 5 2+ = 25 3 ( ) 4 4 Solve = 9 ( ) Solve ( ) 4 25 =

10 Objective 23 Logarithmic Functions Consider an eponential function = a What s the inverse function? There is no algebraic operation to solve for. We must define a new function. = log a Objective 23a Evaluate Logarithmic Functions log 2 8 = log 25 5 = log /6 2 = log 2 2 = log 2 = Which are defined? (Be careful, sometimes ask Which are undefined? ) log /2 log /4 4 log /2 ( 4) log /2 0 86

11 Objective 23b Properties and Graphs of Logarithmic Functions f() = log b, b > 0, b The logarithmic functions are classified into two groups comparable to the eponential functions. Recall Obj 22a = a, a > = a, 0 < a < = log b, b > = log b, 0 < b < Consider f() = log 4 Consider f() = log /4 for an eample of b > for an eample of 0 < b < Pick the -values, find the -values. Pick the -values, find the -values

12 Objective 23c Graphing Logarithmic Functions with Reflections or Translations Don t. Don t. Use Obj 4! Select the graph that best represents the graph of each of the following. f() = log 4 f() = log /4 ( ) Which function best describes the graph shown? Which function best describes the graph shown? f() = log (5/2) () f() = log (5/2) ( ) f() = log (2/5) () f() = log (2/5) ( ) f() = log (5/2) () f() = log (5/2) ( ) f() = log (2/5) () f() = log (2/5) ( ) More Objective 23c Graphing Logarithmic Functions with Translations Don t. Don t. Use Obj 4! 88

13 Select the graph that best represents the graph of each of the following. f() = log 3 ()+2 f() = log /3 (+2) Which function best describes the graph shown? Which function best describes the graph shown? f() = log (5/2) ()+2 f() = log (5/2) () 2 f() = log (2/5) ()+2 f() = log (2/5) () 2 f() = log (5/2) (+2) f() = log (5/2) ( 2) f() = log (2/5) (+2) f() = log (2/5) ( 2) 89

14 Objective 23d Domain of Logarithmic Functions (not b graphing) Give the domain. f() = log b (4 5) f() = 5 log b (3) ( ) + f() = log b 3 f() = log 3 (4 2 ) f() = log 3 ( 2 +4) 90

15 Objective 24 Properties of Logarithmic Functions As used below: a > 0, a, b > 0, b, M > 0, N > 0, > 0, and r represent an real number Definition - Obj 24a means Common Logarithms are logarithms base 0; we write instead of. Natural Logarithms are logarithms base e; we write instead of. Objective 24a Eample Which of the following is equivalent to ln5 =? A) 5 e = B) e = 5 C) 5 = e Properties of Logarithms - Obj 24b Product Rule log b (MN) = Must Note: log b (MN) Must Note: log b (M +N) Quotient Rule ( ) M log b = N ( ) M Must Note: log b N Must Note: log b (M N) Power Rule log b M r = Change-of-Base Formula log b M = log b M = When Base and Result Match log b b = When Result is log b = 9

16 Inverse Function Properties - Obj 24c Recall Obj 2: (f f )() = and (f f)() = a log a M = log a a r = Objective 24c Eamples Solve for if 5 log 5 (3) = 5 Solve for if lne 5 = 3 Objective 24b Eample Which of the following is equivalent to log b ( )? ( ) A) log b C) both A and B B) log b log b D) none is equivalent Objective 24e Eamples Evaluating e or ln on a scientific calculator and rounding the result to 3 decimal places. When rounding a number to 3 decimal places, look at the 4th digit of the decimal. If that number is less than 5, then keep the 3rd digit as it is and drop the remaining decimal digits. If that number is greater than, or equal to 5, then round the 3rd digit up (add to the 3rd digit) and drop the remaining decimal digits. Evaluate, round to 3 decimal places. e 5 ln50 Objective 24be Eample Select all that are correct for log 3 8. Choice: = log8 log3 Choice: = ln3 ln8 Choice:.983 Choice: Objective 24abc Eample Select ALL the correct formulas/statements if b > 0,b, > 0, > 0. (egrade does not warn when a multiple selection problem is left blank.) log(+) = log+log log( ) = log log log = means 0 = log 5 = 0 If log0 4 = 2, then = 3 92

17 Appling Log Properties - Objective ( ) 24d 2 Epand using log properties. log b z ( 2 ) Epand using log properties. log b z(w +3) Which of the following is equivalent to A) 2log b +log b log b z +log b (w+3) B) 2log b +log b log b z log b (w+3) C) A and B are the same log b ( 2 ) log b (z(w +3)) another Objective 24d Eample Write as a single logarithm 2log b log b + log 2 bz A) log b 2 z B) log b 2 z 93

18 another Objective 24d Eample Write as a single logarithm. 2log b (z w) log b w +3log b z +log b log b (+w) another Objective 24d Eample If log b 2 = l and log b 5 = m, epress log b 00 in terms of l and m. another Objective 24d Eample If log b 2 = l and log b 5 = m, epress (log b 4) (log b 25) in terms of l and m. Objective 25 Solving Eponential Equations when we can t obtain the same base. Logarithmic functions are one-to-one. That means: if and onl if Objective 25a Solve. 6 2 = 4 3 Solve = 2 94

19 Objective 25b Round solution to 3 decimal places. Solve. 400e 3+ = Solve. 40e 2 = 20 Solve = Solve = 20 Objective 26 Solving Logarithmic Equations Objective 26a All terms in the equation involve log functions. Solve. log+log( 2) = log(+4) Solve. log 3 (+) log 3 (+7) = log 3 (+5) 95

20 Solve. log 7 ( 8)+log 7 ( 9) = log 7 (23 3) Solve. log 5 (+4) = log 5 (3+8) log 5 (+3) Objective 26b Use the Definition of logarithmic notation. Solve. 5 7log /3 = 5 Solve. log( 2 +2+) = Solve. 6log 4 ( 5) = 3 Solve. log 2 ( 2 +7) = 4 96

21 Objective 27 Solving Equations that involve eponential functions. Solve. e e 2 2 e 4 = 0 Solve. e 3 ( 3 ) + e 3 4 e 6 = 0 Solve. ( 2 ) e e 4 e 2 = 0 Solve. 2e 2 (4+5) 3 e 2 3 (4+5) 2 4 (4+5) 6 = 0 Solve. (3+7) 4 e 4 4 e 4 4(3+7) 3 3 (3+7) 8 = 0 97

22 Solve. (+2) 6 e 3 2 5(+2) 4 2 e 3 = 0 (+2) 8 Solve. 2 ( 3) 5 e 4 2 e 4 ( 3) 3 4 ( 3) 6 = 0 Objective 28 Solving Equations that involve natural logarithmic functions. Solve. 22 ( ( 2ln+2 7 = 0 )) Solve. 44 ( ( ) 8 +8ln = 0 ) Solve. ( 2 2ln ) 2 = 0 Solve. 4 3 (3ln)(43 ) 8 = 0 98

23 Solve. ( 4 3 ln 6ln ) = 0 Solve. (ln) 2 ( ) 3 3 ln 3 4 ln 3 = 0 6 Objective 30 Solve 2 linear equations in 2 unknowns There are 3 possible situations. To solve algebraicall we will use the Multipl one, or both equation if needed, b non-zero numbers so that when the equations are added, one variable is eliminated. 99

24 Solve. 2+5 = 0 Solve. 7 4 = = = 4 Solve. 9 2 = = 2 Solve. 6+0 = = 4 Objective 3 Solve 2 equations in 2 unknowns First tpe: Quadratic and Linear - Algebraic solution To solve algebraicall we will use the Substitute the into the. Solve. = 2 +2 Solve. = = 5 3 = 5 Give the -coordinate(s) onl of an solution(s). If multiple solutions, then enter the values in an order, separated b a semicolon. If there is no solution, then enter: no solution (When there is no solution, do not use an capital letters; do not use an punctuation marks.) 00

25 Second tpe: 2 Various equations - Solve b GRAPHING We will ask: solutions are there? Must know all the basic functions and equations we have graphed and we must know Obj 4 - and r (,) ( h,k) 0

26 Solve the sstem of equations b graphing. How man real solutions are there? = + = 2 Solve the sstem of equations b graphing. How man real solutions are there? = e = 3 Solve the sstem of equations b graphing. How man real solutions are there? = ln ( ) = 4 Copright c 200-present, Annette Blackwelder, all rights are reserved. Reproduction or distribution of these notes is strictl forbidden. 02

is on the graph of y = f 1 (x).

is on the graph of y = f 1 (x). Objective 2 Inverse Functions Illustrate the idea of inverse functions. f() = 2 + f() = Two one-to-one functions are inverses of each other if (f g)() = of g, and (g f)() = for all in the domain of f.