1. 6.1 The Composition of Functions f o g - composition of the function f with g is is defined by the equation (f o g)(x) = f (g(x)). The domain is the set of all numbers x in the domain of g such that g(x) is in the domain of f. f (x) = 3x – 4 and g(x) = x 2 + 6 (f o g)(x) = f (g(x)) = 3g(x) – 4 = 3(x 2 + 6) – 4 = 3x 2 + 18 – 4 = 3x 2 + 14 (g o f)(x) = g(f (x)) = (f (x)) 2 + 6 = (3x – 4) 2 + 6 = 9x 2 – 24x + 16 + 6 = 9x 2 – 24x + 22 What is you already know (f o g)(x) and you want to find it’s components f & g? Let (f o g)(x) = 1(x + 1) Find f(x) & g(x) (f o g)(x) appears to be a reciprocal function of (x + 1) so, g(x) = (x + 1) f(x) = 1/x

2. Finding the Domain of (f o g)(x) f o g - composition of the function f with g is is defined by the equation (f o g)(x) = f (g(x)). The domain is the set of all numbers x in the domain of g such that g(x) is in the domain of f. Find the domain of if f(x) = 1/(x+2) and g(x) = 4/(x – 1) The domain of g : {x | x ≠ 1} The domain of f : {x | x ≠ -2} so we must exclude any g(x) that would yield -2 for f o g When is g(x) = -2 ? 4/(x + 1) = -2 =>-2(x + 1) = 4 => x +1 = -2 => x = -1 So, the domain of is : {x | x ≠ 1, x ≠ -1} Note that you can also find the domain after finding f(g(x)). (f o g)(x)

3. 6.2 Inverse Functions/Horizontal Line Test Function – Every element x in the domain corresponds to one and only one element y One-to-one function – each x value corresponds to one and only one y value and each y value corresponds to one and only one x value if x 1 and x 2 are different inputs of f then f(x 1 ) ≠ f(x 2 ) A function f has an inverse that is a function, f –1, if it is one-to-one. (There will be no horizontal line that intersects the graph of the function f at more than one point) f(x) = x 2 +3x-1 NO Inverse Function y x f(x) = x + 4 YES this has an inverse Function

4. Using Diagrams to determine If a Function or Inverse Exists Yes! - InverseNo Inverse

5. Graphing the Inverse of a Function The domain of f is the range of f -1 The range of f is the domain of f -1 The graphs of f and f -1 are symmetric reflections of each other across the line y = x To graph the inverse (f --1 ) reverse each (x, y) point on f, graphing (y, x). f --1 = {(y, x) | (x, y) belongs to f}

6. Verifying Inverse Functions 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. Then the function g is the inverse of the function f denoted by f -1 and the function f is the inverse of the function g denoted by g -1 The domain of f is equal to the range of f -1, and vice versa. (x,y) in f => (y, x) in f --1 Examples: Verifying inverses (Are f & g inverses?) f (x) = 5x and g(x) = x/5. f(x)= 3x + 2 g(x) = x - 2 3 f (g(x)) = 5(g(x)) = 5 x = x f (g(x)) = 3(g(x))+2 5 = 3 x-2 + 2 g( f (x)) = f(x) = 5x = x 3 5 5 = x – 2 + 2 = x f (g(x)) = x and g( f (x)) = x g(f(x) = f(x) – 2 = 3x + 2 – 2 = x 3 3 Thus they are inverses. Thus they are inverses f(x) = 5x f -1 (x)=x/5 f(x ) = 3x + 2 f -1 (x) = (x-2)/3

7. Example: Find the inverse of f (x) = 7x – 5. Step 1: Replace f (x) by y: y = 7x – 5 Step 2: Interchange x and y : x = 7y – 5. Step 3: Solve for y. : x + 5 = 7y x + 5 = y 7 Step 4 Replace y by f -1 (x). x + 5 7 f -1 (x) = How to Find the Inverse of a Function

8. Find the Range of a Function (an algebraic method) f(x) = 2x + 1 x - 1 Find the domain and range of f(x) The domain of f is: {x | x ≠ 1} X cannot be 1 since it would make the denominator 0. To find the range first, find the inverse, f -1 Interchange x & y in f(x) then solve for y. x = 2y + 1 => x(y-1) = 2y + 1 => xy – x = 2y + 1 y -1 => xy – 2y = x + 1 => y(x -2) = x + 1 => y = (x + 1) (x – 2) y in this case represents f -1 The domain here is: {x | x ≠ 2} Therefore, the range of f(x) is: {x | x ≠ 2}

9. Graphing the Inverse of a Function Exercise y x 1.Draw any function (f) that passes the horizontal line test. 2. To graph the inverse (f --1 ) reverse each (x, y) point on f, graphing (y, x). Create your own Example:

10. 6.3 Exponential Functions Exponential function – any function whose equation contains a variable in the exponent. [measures rapid increase or decrease (Example: epidemic growth)] f(x) = b x f – exponential functionb - constant base (b > 0, b  1) x = any real number f(x) = 2 x g(x) = 10 x h(x) = 3x+1 3 2+1 = 3 3 = 273 2 = 92 3 1+1 = 3 2 = 93 1 = 31 3 0+1 = 3 1 = 33 0 = 10 3 -1+1 = 3 0 = 13 -1 = 1/3 3 -2+1 = 3 -1 = 1/33 -2 = 1/9-2 g(x) = 3 x+1 f (x) = 3 x x g(x) = 3 x+1 (0, 1) (-1, 1) 123456 -5-4-3-2 Graphing Exponential Functions: Shift up c units f(x) = b x + c Shift down c units f(x) = b x – c Shift left c units f(x) = b x+c Shift right c units f(x) = b x-c

11. All properties above are the same Except #4 which is ‘decreasing’

12. The Natural Base e An irrational number, symbolized by the letter e, appears as the base in many applied exponential functions. This irrational number is approximately equal to 2.72. More accurately, The number e is called the natural base. The function f (x) = e x is called the natural exponential function. f (x) = e x f (x) = 2 x f (x) = 3 x (0, 1) (1, 2) 1 2 3 4 (1, e) (1, 3) P. 429 in your book shows you more about where e comes from.

13. Solving Exponential Equations 2 5 = 32, so 2 3x-1 = 2 5  3x – 1 = 5  3x = 6  X = 2 e 2x-1 = e -4x e 3x  e 2x-1 = e -7x  2x – 1 = -7x  -1 = -9x  X = 1/9

14. Application: Depreciation: The price, p, in dollars of a Honda Civic DX Sedan that is x years old is given by: P(x) = 16,630 (0.90) x (a)How much does a 3-year old Civic DX Sedan cost? (b)How much does a 9-year old Civic DX Sedan cost? Application: Drug Medication: The function D(h) = 5e -0.4h can be used to find the number of milligrams D of a certain drug that is in a patient’s bloodstream h hours ofter the drug has been adminstered. (a)How many milligrams will be present after 1 hour? (b)How many milligrams will be present after 6 hours?

15. 6.4 & 6.5 Logarithmic Functions A logarithm is an exponent such that for b > 0, b  1 and x > 0 y = log b x if and only if b y = x Logarithmic equations Corresponding exponential forms 1)2 = log 5 x1) 5 2 = x 2)3 = log b 642) b 3 = 64 3)log 3 7 = y3) 3 y = 7 4)y = log e 94) e y = 9 log 25 5 = 1/2 because 25 1/2 = 5.25 to what power is 5?log 25 5 log 3 9 = 2 because 3 2 = 9.3 to what power is 9?log 3 9 log 2 16 = 4 because 2 4 = 16.2 to what power is 16?log 2 16 Logarithmic Expression Evaluated Question Needed for Evaluation Evaluate the Logarithmic Expression

16. Logarithmic Properties Log b b = 11 is the exponent to which b must be raised to obtain b. (b 1 = b). Log b 1 = 00 is the exponent to which b must be raised to obtain 1. (b 0 = 1). log b b x = xThe logarithm with base b of b raised to a power equals that power. b log b x = xb raised to the logarithm with base b of a number equals that number. Graphs of f (x) = 2 x and g(x) = log 2 x [Logarithm is the inverse of the exponential function] 4 2 8211/21/4f (x) = 2 x 310-2x 2 4 310-2g(x) = log 2 x 8211/21/4x Reverse coordinates. -2 6 2345 5 4 3 2 -2 6 f (x) = 2 x f (x) = log 2 x y = x Properties of f(x) = log b x Domain = (0, ∞) Range = (-∞, +∞) X intercept = 1 ; No y-intercept Vertical asymptote on y-axis Decreasing on 0 1 Contains points: (1, 0), (b, 1), (1/b, -1) Graph is smooth and continuous

17. Common Logs and Natural Logs A logarithm with a base of 10 is a ‘common log’ log 10 1000 = ______ because 10 3 = 1000 If a log is written with no base it is assumed to be 10. log 1000 = log 10 1000 = 3 3 A logarithm with a base of e is a ‘natural log’ log e 1 = ______ because e 0 = 1 If a log is written as ‘ln’ instead of ‘log’ it is a natural log ln 1 = log e 1 = 0 0

18. More Properties of Logarithms Basic Properties Log b b = 11 is the exponent to which b must be raised to obtain b. (b 1 = b). Log b 1 = 00 is the exponent to which b must be raised to obtain 1. (b 0 = 1). Inverse Properties log b b x = xThe logarithm with base b of b raised to a power equals that power. b log b x = xb raised to the logarithm with base b of a number equals that number. log b (MN) = log b M + log b N Product Rule Quotient Rule log b M = log b M - log b N N log b M = p log b M Power Rule p For M>0 and N > 0

19. Logarithmic Property Practice log b (MN) = log b M + log b N 1)log 3 (27 81) = 2) log (100x) = 3) Ln (7x) = log b M = log b M - log b N N log b M = p log b M p 1) log 8 23 x 2) Ln e 5 11 1)log 5 7 4 2)Log (4x) 5 3)Ln x 2 = 4)Ln x = = = = = Product Rule Power Rule Quotient Rule

20. Expanding Logarithms log b (MN) = log b M + log b N log b M = log b M - log b N N log b M = p log b M p 2) log 6 3 x 36y 4 1) Log b (x 2 y ) 3) log 5 x 25y 3 4)log 2 5x 2 3

21. Condensing Logarithms log b (MN) = log b M + log b N log b M = log b M - log b N N log b M = p log b M p 1)log 4 2 + log 4 324) 2 ln x + ln (x + 1) 2)Log 25 + log 45) 2 log (x – 3) – log x 3)Log (7x + 6) – log x6) ¼ log b x – 2 log b 5 – 10 log b y Note: Logarithm coefficients Must be 1 to condense. (Use power rule 1 st )

22. The Change-of-Base Property Example: Evaluate log 3 7 Most calculators only use: Common Log [LOG] (base 10) Natural Log [LN] (base e) It is necessary to use the change Of base property to convert to A base the calculator can use.

23. 6.6 Using Natural Logarithms to Solve Exponential Equations Step 1: Isolate the exponential expression. Step 2: Take the natural logarithm on both sides of the equation. Step 3: Simplify using one of the following properties: ln b x = x ln b or ln e x = x. Step 4: Solve for the variable using proper algebraic rules. 5 4x – 7 – 3 = 10 log 4 (x + 3) = 2. 3 x+2 -7 = 27 Examples (Solve for x): log 2 (3x-1) = 18

24. More Equations to Try

25. 6.7 Compound Interest If a principal of P dollars is borrowed for a period of t years at a perannum interest rate r, the interest I charged is: I = Prt Annually -> Once per yearSemiannually -> Twice per year Quarterly -> Four times per yearMonthly -> 12 times per year Daily -> 365 times per year Example: A credit union pays interest of 8% per annum compounded quarterly on a Certain savings plan. If $1000 is deposited in such a plan and the interest is Left to accumulate, how much is in the account after 1 year? I = (1000)(.08)(1/4) = $20 => New principal now is: $1020 I = (1020)(.08)(1/4) = $20.40 => New principal now is: $1040.40 I = (1040.40)(.08)(1/4) = $20.81 => New principal now is: $1061.21 I = (1061.21)(.08)(1/4) = $21.22 => New principal now is: $1082.43 Compound Interest Formula: The amount, A, after t years due to a principal P Invested at an annual interest rate r compounded n times per year is: A = P  (1 + r/n) nt

26. 6.8 Exponential Growth & Decay

27. Example: Bacterial Growth A colony of bacteria that grows according to the law of uninhibited growth Is modeled by the function N(t) = 100e 0.045t where N is measured in grams and t is measured in days. (a)Determine the initial amount of bacteria N 0 (initial amount) occurs when t = 0. so, N(0) = 100e 0.045(0) = 100 grams (b)What is the population after 5 days? N(5) = 125.2 grams (c)How long will it take for the population to reach 140 grams? 140 = 100e 0.045t => t = 7.5 days (d) What is the doubling time for the population? 200 = 100e 0.045t => t = 15.4 days