Exponential functions have a variable in theExponent and a numerical base. Ex.

Not to be confused with power functions which Have a variable base. Ex.

y =2x, y=3(4)x

y =x4 , y=2x5

A function that can be expressed in the form

above is called an Exponential Function. Exponential Functions with positive

values of x are increasing, one-to-one functions.

For f(x) = bx the graph has a y-intercept at (0,1) (initial value) and passes through (1,b).

The value of b determines the steepness of the curve.

The function is neither even nor odd. There is no symmetry.

There is no local extrema.

( ) , 0, 1xf x a b a b

f (x) =a⋅bx

The domain: The range: End Behavior:

The y-intercept is The horizontal

asymptote:

, . 0, .

limx→ −∞

=0, limx→∞

=∞

0.y 0,1 .

There is no x-intercept.There is no x-intercept. There are no vertical asymptotes.There are no vertical asymptotes. This is an increasing, continuous This is an increasing, continuous function.function. Initial value is a.Initial value is a.

How would you graph

( ) 3 ?xf x

( ) 6 ?xf x

Domain:

Range:

Y-intercept:

, 0,

0,1Hor.Asy:

0y How would you graphHow would you graph

Naming a, b ( ) 3 ?xf x

f (x) 1• 3x

a, (initial

value)=1:

b, growth rate =

3

Rewritten:Rewritten: In

form y =abx

Recall that if reflection of about the y-axis.

Thus, if

1( )

x

f xb

→ f (x) =b−x

( )f x

f (x) =2−x, decreasingInitial value: (0,1)

Rate of decay: ½

H.A.:

0y

Notice that the reflection is decreasing, or decaying at a rate of 1/3 and the end behavior is:

limx→∞

f x→ 0. limx→−∞

f x→ ∞.

1( ) ?

3

x

f x

How would you graphHow would you graph

If b>1, then•f is an increasing function,• andlim ( ) 0

xf x

lim ( )

xf x

If 0<b<1, then•f is a decreasing function,• andlim ( )

xf x

lim ( ) 0

xf x

If a =1, (0,1) is the y interceptIf a > 1, then (0,a) is the y intercept

y =4(3)x

a = 4, b = 3, (0, 4) is the initial valueWhich means when x = 0, y = 4

Given: (0,6), (1, 12), (2, 24), (3, 48)

Hint: find a first, then substitute onepoint into to solve for b. y =abx

y =abx

Given: (0,6), (1, 12), (2, 24), (3, 48)a = 6 to solve for b:

SOLUTION:

y =6• bx

24 =6 • b2

4 =b2

b=2y =6(2)x

• Exponential graphs, like other functions we have •studied, can be stretched/shrunk, reflected and• translated.• It is important to maintain the same base as you• analyze the transformations.

( ) 2 3xg x Vertical shift up 3:

( ) 3(2 ) 1xg x

Reflect in x-axisVertical stretch 3Vertical shift down 1

1(2) 1xy 212 (3) 3xy

Domain:

Range:

HA

Domain

:Range:

H.Asym:

Inc/decreasing?

Inc/decreasing ?

1(2) 1xy Reflect about the x-axis.Horizontal shift right 1.Vertical shift up 1.

212 (3) 3xy

Vertical shrink ½ .Horizontal shift left 2.

Vertical shift down 3.

Domain:

Range:

HA

Domain

:Range:

Y-intercept:

H.Asym:

, .

,1 . 1.y

decreasing

, .

3, . 3.y

320, .

increasing

( ) xf x e

lim 0x

xe

lim x

xe

( ) 3 2xf x e 2 2xf x e 1xf x e

Domain:

Range:

Y-intercept:

H.A.:

Domain:

Range:

Y-intercept:

H.A.:

Domain:

Range:

Y-intercept:

H.A.:

, , ,

2, 0,5

2y

, 1 0, 2

1y

2,

0,9.389

2y

TransformationsTransformations

Vertical stretch 3.Vertical shift up 2.

Reflect @ x-axis.Vertical shift down 1.

Horizontal shift left 2.Vertical shift up 2

Inc/dec? increasing

Concavity?up

Inc/dec? Inc/dec?decreasing

Concavity?down

increasing

Concavity? up

Exponential function: y = abx

Growth rate: A = a(1+r)x

Decay rate: A = a(1-r)x

If b>1 it’s a growth…..What is the rate of growth?

If b<1, it’s a decay…What is the rate of decay?

f (x) 31.05x

g(x) 2.5.98t

Growth rate: A = a(1+r)t

Decay rate: A = a(1-r)t

What is the rate of growth?1.05 – 1 = .05 = 5%

What is the rate of decay?1 - .98 = .02 = 2%

f (x) 31.05x

g(x) =2.5(.98)t

Growth rate: A = a(1+r)t

Decay rate: A = a(1-r)t

What is the rate of growth?1.05 – 1 = .05 = 5%

What is the rate of decay?1 - .98 = .02 = 2%

See page 296 #13,14,20, 31

f (x) 31.05x

g(x) =2.5(.98)t

Radioactive decay or growth, (half life isan example)

Growth model: (b > 1), t may have a coefficient

A = a0 (b)t

Decay model: (0<b<1)Half life: (K is the half life of the substance…)

A = a0 (.5) t/k t is the time we are calculating k is the half life of the substance

The half life of a substance is 20 days.Express the amount of the substance remaining as a function of time, t, whenthe initial amount is 30 grams

Use the function to determine how manygrams remain after 40 days.

The half life of a substance is 20 days.Express the amount of the substance remaining as a function of time, t, whenthe initial amount is 30 grams

Use the function to determine how manygrams remain after 40 days.

y =30(.5)t20

y =30(.5)4020 =7.5

When will the substance remainingbe 5 grams?Graph the function and check the Table of values

y =30(.5)t20

Now try page 297 31

Do Now:Page 298 # 55

“logistic growth”

C is max value

a is initial value, b is growth

Logistic function

y =c

1+abx

A problem that seems reasonable:

Under favorable conditions, a single cell of the bacterium Escherichia coli divides into two about every 20 minutes. If the same rate of division is maintained for 10 days, how many organisms will be produced from a single cell?

Solution:

10 days = 720 20-minute periods

There will be 1 ∙ 2^720 ≈ 5.5 ∙ 10^216 bacteria after 10 days.

Makes sense…

…until you consider that there are probably fewer than 10^80 atoms in the entire universe.

Real world Bizarro world

Why didn’t they tell us the truth? Most of those classical “exponential growth” problems should have been “logistic growth” problems!

Exponential Logistic

The logistics function is like capping off an Exponential function.

f (x) =c

1+abx

Where c is the limited growth and a is determined By the initialValue. It is bounded by y=0 and y=c

Do page 288 # 41,50

see page 297 # 24Find the initial value first, (0,12)

Cross multiply and solve for a

12 =60

1+ab0

see page 297 # 24 Now solve for b:

24 =60

1+ 4b1

see page 297 # 24 Now solve for b:

24 =60

1+ 4b1

24 +96b=60b=.375

y =60

1+ 4(.375)x

f (x) =c

1+abx

Now try 26 and 28 from page 297

f (x) =c

1+abx

You may use the calculator to find the Point of symmetry.See page 288, # 49 to find it.

The point of symmetry for a logistics function is at the point:

(x,.5c)

Common (base 10) and natural logarithms (base e):“a logarithm is an exponent”

Exponentials and logarithms are inverses

log 10 = 1, log 100 = 2, log 1 = 0

ln e = 1, ln e2 = 2, ln 1 = 0

log (-5) does not exist!

log 100

log2 2

log5 25

log5 1

Examples:

Common and natural logarithms:“a logarithm is an exponent”

log 100 =log10 (102 )

12 =101 =1

log2 2 =log2 212 =

12

log5 25 =log5 52 =2

log51=0

Common and natural logarithms:“a logarithm is an exponent”

log6 365

log2

1

8

log5 1256

Common and natural logarithms:“a logarithm is an exponent”

log6 365 =log6 625 =log6 625 =

25

log218

=log21

23=2

−32 =−

32

log5 1256 =log5 536 =log5 536 =

12

Practice page 30812-24 evens

New example:

Now try 34 and 36

Hw: page 308 1-35 odds

25x = 5

Express each side as a base 5 number:

Equations that contain one or more exponential expressions are called exponential equations.

Steps to solving some exponential equations:

1. Express both sides in terms of same base.2. When bases are the same, exponents are equal. i.e.:

52( )

x =512 → 52x =5.5

2x=.5→ x=.25

DO NOW: Get like bases and solve:

3 14 32x

Get like bases:3 14 32x 3 12 52 2

x

6 2 52 2x 6x 2 5

6 3x 1

2x

Set exponents equal:

Solve and check!

4x2+4x =4−3

Isolate the base & solve

2 • 5x4 =250

Isolate the base solve

2 • 5x4 =250

5x

4 =125 5x

4 =53

x4

=3

x=12

Isolate the base & solve

2 • 5x4 =250

Isolate the base & solve

2 • 5x4 =250

Isolate the base “log both sides” & solve

5x

4 =125

log 5x

4 =log125

“log both sides” & solve

5x

4 =125

log 5x

4 =log125x

4log 5 log125

x4

log125log5

x

4=3→ x=12

Exponential EquationsExponential Equations

Sometimes it may be helpful to Sometimes it may be helpful to factor the equation to solve:factor the equation to solve:

2 4 0xx e

2 4 0x 0xe 2 4x

2x There is no value of x forwhich is equal to 0.

xe

or

2 4x xxe e

2 4 0x xxe e

2 3 34x xx e e

Try this…….

Exponential EquationsExponential Equations

2 3 34x xx e e

2 3 34 0x xx e e

3 2 4 0xe x 3 0xe or

2 4 0x 2 4x

2x

INTEREST FORMULA

amount at the end

Principal(amount at

start)

annual interest rate(as a decimal)

A =P 1+ r( )t time

(in years)

You invest 1200 into a 12 month cdat 2% interest. How much is it worthat the end of the year?

You invest 1200 into a 12 month cdat 2% interest. How much is it worthat the end of the year?

p =1200(1+.02)1

p=1224

COMPOUND INTEREST FORMULA

amount at the end

Principal(amount at

start)

annual interest rate(as a decimal)

A =P 1+rn

⎛

⎝⎜

⎞

⎠⎟

nttime

(in years)

number of times per year that interest in compounded

You invest 1200 into a 12 month cdat 2% compounded quarterly. How much is it worthat the end of the year?

You invest 1200 into a 12 month cdat 2% compounded quarterly. How much is it worthat the end of the year?

Find the amount of an investment of $500 Compounded quarterly for two years at 8 % annual rate.

nt

n

rPA

1

500

.08

4

4 (2)

83.585$A

.

Find the amount that results from $500 invested at 8% compounded quarterly after a period of 2 years.

A =A0 (1+rn)nt

Solve for A0:

A =600,r =6% n=12 t =4

A =A0 (1+rn)nt

Solve for A0:

A =600,r =6% n=12 t =4

600 =A0 1+.0612

⎛

⎝⎜

⎞

⎠⎟12• 4

A =A0 (1+rn)nt

A =600,r =6% n=12 t =4

600 =A0 1+.0612

⎛

⎝⎜

⎞

⎠⎟12• 4

600 =A0 (1.005)48

A0 =600

1.27049≈472.26

What about continously?What does that mean?Things that are continuous: Population, medicine, compound continuous

Quarterly = 4 times a yearMonthly = 12Semi-annually = 2Daily = 360

The number e

•The letter e is the initial of the last name of Leonhard Euler (1701-1783) who introduced the notation.

• Since has special calculus properties that simplify many calculations, it is the natural base of exponential functions.

•The value of e is defined as the number that the expression approaches as n approaches infinity.

• The value of e to 16 decimal places is 2.7182818284590452.

• The function is called the Natural Exponential Function

11

n

n

( ) xf x e

( ) xf x e

11

n

n

Try it for n = 1000Try it for n = 10,000

p =pert

Compound continously:

Calculate:

How much money will you have if you invest 500 at an annual rate of 8 %, compounded continuously for two years.

Calculate:

How much money will you have if you invest 500 at an annual rate of 8 %, compounded continuously for two years.

p =500e.08(2)

p=586.76

Do Now: copy the rules on page 310 in the green box.

Logarithmic Properties…

Then try page 317 1,3,5

See “Change of base formula” on page 313

Do # 33 on page 317