6.1 The Composition of Functionsf 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) = x2 + 6

(f o g)(x) = f (g(x)) = 3g(x) – 4 = 3(x2 + 6) – 4 = 3x2 + 18 – 4 = 3x2 + 14

(g o f)(x) = g(f (x)) = (f (x))2 + 6= (3x – 4)2 + 6= 9x2 – 24x + 16 + 6= 9x2 – 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

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)

(f o g)(x)

6.2 Inverse Functions/Horizontal Line TestFunction – 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 x1 and x2 are different inputs of f then f(x1) ≠ f(x2)

A function f has an inverse that is a function, f –1, if it is one-to-one. (There will be nohorizontal line that intersects the graph of the function f at more than one point)

f(x) = x2+3x-1

NO Inverse Function

y

x

f(x) = x + 4

YES this has an inverse Function

Using Diagrams to determineIf a Function or Inverse Exists

Yes! - Inverse No Inverse

No Inverse

Graphing the Inverse of a FunctionThe 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}

Verifying Inverse FunctionsIf 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 3f (g(x)) = 5(g(x)) = 5 x = x f (g(x)) = 3(g(x))+2 5 = 3 x-2 + 2g( 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 3Thus they are inverses. Thus they are inversesf(x) = 5x f-1(x)=x/5 f(x ) = 3x + 2 f-1(x) = (x-2)/3

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 = y7

Step 4 Replace y by f -1(x).

x + 57

f -1(x) =

How to Find the Inverse of a Function

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}

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:

6.3 Exponential FunctionsExponential function – any function whose equation contains a variable in the exponent. [measures rapid increase or decrease (Example: epidemic growth)]

f(x) = bx

f – exponential function b - constant base (b > 0, b 1) x = any real number

f(x) = 2x g(x) = 10x h(x) = 3x+1

32+1 = 33 = 2732 = 92

31+1 = 32 = 931 = 31

30+1 = 31 = 330 = 10

3-1+1 = 30 = 13-1 = 1/3-1

3-2+1 = 3-1 = 1/33-2 = 1/9-2

g(x) = 3x+1f (x) = 3xxf (x) = 3x

g(x) = 3x+1

(0, 1)(-1, 1)

1 2 3 4 5 6-5 -4 -3 -2 -1

Graphing Exponential Functions:Shift up c unitsf(x) = bx + c

Shift down c unitsf(x) = bx – c

Shift left c unitsf(x) = bx+c

Shift right c unitsf(x) = bx-c

All properties above are the sameExcept #4 which is ‘decreasing’

The Natural Base eAn 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) = ex is called the natural exponential function.

-1

f (x) = ex

f (x) = 2x

f (x) = 3x

(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.

3 -1Solve: 2 32x 42 13

1Solve: x x

xe e

e

Solving Exponential Equations

25 = 32, so

23x-1 = 25

3x – 1 = 5

3x = 6

X = 2

e2x-1 = e-4x

e3x

e2x-1 = e-7x

2x – 1 = -7x

-1 = -9x

X = 1/9

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?

6.4 & 6.5 Logarithmic FunctionsA logarithm is an exponent such that for b > 0, b 1 and x > 0 y = logb x if and only if by = x

Logarithmic equations Corresponding exponential forms1) 2 = log5 x 1) 52 = x2) 3 = logb 64 2) b3 = 643) log3 7 = y 3) 3y = 74) y = loge 9 4) ey = 9

log25 5 = 1/2 because 251/2 = 5.25 to what power is 5?log25 5

log3 9 = 2 because 32 = 9.3 to what power is 9?log3 9

log2 16 = 4 because 24 = 16.2 to what power is 16?log2 16

Logarithmic Expression Evaluated

Question Needed for Evaluation

Evaluate the

Logarithmic Expression

Logarithmic Properties

Logb b = 1 1 is the exponent to which b must be raised to obtain b. (b1 = b).

Logb 1 = 0 0 is the exponent to which b must be raised to obtain 1. (b0 = 1).

logb bx = x The logarithm with base b of b raised to a power equals that power.b logb x = x b raised to the logarithm with base b of a number equals that number.

Graphs of f (x) = 2x and g(x) = log2 x [Logarithm is the inverse of the exponential

function]

4

2

8211/21/4f (x) = 2x

310-1-2x

2

4

310-1-2g(x) = log2 x

8211/21/4x

Reverse coordinates.

-2 -1

6

2 3 4 5

5

4

3

2

-1-2

6

f (x) = 2x

f (x) = log2 x

y = x

Properties of f(x) = logb x

•Domain = (0, ∞)•Range = (-∞, +∞)•X intercept = 1 ; No y-intercept•Vertical asymptote on y-axis•Decreasing on 0<b<1; increasing if b>1•Contains points: (1, 0), (b, 1), (1/b, -1)•Graph is smooth and continuous

Common Logs and Natural Logs

A logarithm with a base of 10 is a ‘common log’log10 1000 = ______ because 103 = 1000

If a log is written with no base it is assumed to be 10.

log 1000 = log10 1000 = 3

3

A logarithm with a base of e is a ‘natural log’loge 1 = ______ because e0 = 1

If a log is written as ‘ln’ instead of ‘log’ it is a natural log

ln 1 = loge 1 = 0

0

More Properties of LogarithmsBasic PropertiesLogb b = 1 1 is the exponent to which b must be raised to obtain b. (b1 = b).

Logb 1 = 0 0 is the exponent to which b must be raised to obtain 1. (b0 = 1).Inverse Properties

logb bx = x The logarithm with base b of b raised to a power equals that power.b logb x = x b raised to the logarithm with base b of a number equals that number.

logb(MN) = logb M + logb N Product Rule

Quotient Rule

logb M = logb M - logb N N

logb M = p logb M Power Rule

p

For M>0 and N > 0

Logarithmic Property Practice

logb(MN) = logb M + logb N

1) log3 (27 • 81) =

2) log (100x) =

3) Ln (7x) =

logb M = logb M - logb N N

logb M = p logb M p

1) log8 23 x

2) Ln e5

11

1) log5 74

2) Log (4x)5

3) Ln x2 =4) Ln x =

=

=

=

=

Product Rule

Power Rule

Quotient Rule

Expanding Logarithmslogb(MN) = logb M + logb N

logb M = logb M - logb N N

logb M = p logb M p

2) log6 3 x36y4

1) Logb (x2 y ) 3) log5 x25y3

4) log2 5x2

3

Condensing Logarithmslogb(MN) = logb M + logb N

logb M = logb M - logb N N

logb M = p logb M p

1) log4 2 + log4 32 4) 2 ln x + ln (x + 1)

2) Log 25 + log 4 5) 2 log (x – 3) – log x

3) Log (7x + 6) – log x 6) ¼ logb x – 2 logb 5 – 10 logb y

Note: Logarithm coefficientsMust be 1 to condense. (Use power rule 1st)

The Change-of-Base Property

Example: Evaluate log3 7

Most calculators only use:

• Common Log [LOG] (base 10)• Natural Log [LN] (base e)

It is necessary to use the changeOf base property to convert toA base the calculator can use.

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 bx = x ln b or ln ex = x.

Step 4: Solve for the variable using proper algebraic rules.

54x – 7 – 3 = 10log4(x + 3) = 2. 3x+2-7 = 27

Examples (Solve for x):

log 2 (3x-1) = 18

More Equations to Try

3 3Solve: log 4 2log x

2 2Solve: log 2 log 1 1x x

Solve: 3 7x

Solve: 5 2 3x

1 2 3Solve: 2 5x x

Solve: 9 3 6 0x x

6.7 Compound InterestIf 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 = PrtAnnually -> Once per year Semiannually -> Twice per yearQuarterly -> Four times per year Monthly -> 12 times per yearDaily -> 365 times per year

Example: A credit union pays interest of 8% per annum compounded quarterly on aCertain 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: $1020I = (1020)(.08)(1/4) = $20.40 => New principal now is: $1040.40I = (1040.40)(.08)(1/4) = $20.81 => New principal now is: $1061.21I = (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 PInvested at an annual interest rate r compounded n times per year is:

A = P (1 + r/n)nt

6.8 Exponential Growth & Decay

Example: Bacterial Growth

A colony of bacteria that grows according to the law of uninhibited growthIs modeled by the function N(t) = 100e0.045t where N is measured in grams and t is measured in days.

(a)Determine the initial amount of bacteria N0 (initial amount) occurs when t = 0. so, N(0) = 100e0.045(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 = 100e0.045t => t = 7.5 days

(d) What is the doubling time for the population? 200 = 100e0.045t => t = 15.4 days