introduction we are going to look at exponential functions we will learn about a new ‘special’...

Introduction

• We are going to look at exponential functions

• We will learn about a new ‘special’ number in Mathematics

• We will see how this number can be used in practical problems…

The Exponential and Log Functions

3A

Imagine you have £100 in a bank account

Imagine your interest rate for the year is 100%

You will receive 100% interest in one lump at the end of the year, so you will now have £200 in the bank

However, you are offered a possible alternative way of being paid

Your bank manager says, ‘If you like, you can have your 100% interest split into two 50% payments, one made halfway through the year, and one made at the end’

How much money will you have at the end of the year, doing it this way (and what would be the quickest calculation to work that out?)

£100 x 1.52

= £225

Investigate further. What would happen if you split the interest into 4, or 10, or 100 smaller bits etc…


3A

£100e£100 x (1 + 1/n)n100/nn£100

£271.81£100 x 1.0001100000.01%10,000£100

£271.69£100 x 1.00110000.1%1,000£100

£270.48£100 x 1.011001%100£100

£269.16£100 x 1.02502%50£100

£265.33£100 x 1.05205%20£100

£259.37£100 x 1.11010%10£100

£256.58£100 x 1.125812.5%8£100

£244.14£100 x 1.25425%4£100

£225£100 x 1.5250%2£100

£200£100 x 2100%1£100

Total (2dp)SumInterest Each

PaymentPayments

Start Amount

11

n

en

The larger the value of n, the better the accuracy of e…The value of e is irrational, like π…It also has another interesting property…

(2.718281828459…)


3A

Gradient Functions

You have already learnt about differentiation, that differentiating a graph function gives the gradient function…

We can plot the gradient function on the graph itself…

y = x2

So dy/dx = 2x

y = x2

y = 2x

The Gradient at this point…

… is this value here!


3A

Gradient Functions



y = x3

So dy/dx = 3x2

y = x3y = 3x2

The Gradient at this point…

… is this value here!

And the Gradient is the same

here!


3A

Gradient Functions



y = 2x

dy/dx = 2xln2

At this stage, you do not need to know where this comes

from…

y = 2x

y = 2xln2


3A

Gradient Functions



y = 3x

dy/dx = 3xln3

At this stage, you do not need to know where this comes

from…

y = 3x

y = 3xln3


3A

y = 3xy = 3xln3

y = 2x y = 2xln2

What has happened from the first graph to the second?

The lines have crossed…

Therefore the must be a value between 2 and 3 where the lines are equal…


3A

Gradient Functions

If we plot a graph of ex, its gradient function is the same graph!

This leads to an interesting conclusion…

If y = ex

Then dy/dx = ex as well!

y = ex

y = ex


• The Graph of ex is used rather than 2x or 3x because e is the increase when growth is continuous (ie growth is always happening, rather than in ‘chunks’)

• The majority of growth follows this format, bacterial growth is continuous for example.

• Transforming the graph can also allow graphs to show decreases, such as radioactive decay and depreciation

3A

You need to be able to sketch transformations of the graph y = ex

y = ex

y = 2ex

3A

y = ex

(0,1)f(x)

2f(x)

y = 2ex

(0,2)


(For the same set of inputs (x), the

outputs (y) double)


y = ex

y = ex + 2

3A

y = ex

(0,1)f(x)

f(x) + 2

y = ex + 2

(0,3)


(For the same set of inputs (x), the

outputs (y) increase by 2)


y = ex

y = -ex

3A

y = ex

(0,1)

f(x)

-f(x)

y = -ex

(0,-1)


(For the same set of inputs (x), the outputs (y) ‘swap

signs’


y = ex

y = e2x

3A

y = ex

(0,1)

f(x)

f(2x)

y = e2x


(The same set of outputs (y) for half

the inputs (x))


y = ex

y = ex + 1

3A

y = ex

(0,1)

f(x)

f(x + 1)

y = ex + 1


(The same set of outputs (y) for

inputs (x) one less than before…)

(0,e)

We can work out the y-intercept by substituting in x = 0

This gives us e1 = e


y = ex

y = e-x

3A

y = ex

(0,1)

f(x)

f(-x)

y = e-x


(The same set of outputs (y) for inputs with the opposite sign…

(0,1)


Sketch the graph of:

y = 10e-x

3A

y = ex

The graph of e-x, but with y values 10 times bigger…

y = e-

x

The Exponential and Log Functionsy =

10e-x

(0, 1)

(0, 10)



y = 3 + 4e0.5x

3A

y = ex

The graph of e0.5x, but with y values 4 times bigger with 3

added on at the end…

(0, 1)

(0, 7)

y = e0.5x

y = 4e0.5x

y = 3 + 4e0.5x (0, 4)



You need to be able to used the Exponential and Log

Functions to solve problems…

The Price of a used car is given by the formula:

a) Calculate the value of the car when it is new The new price implies t, the time, is 0… Substitute t = 0 into the formula…

3A

P = 16000e

- t 10

P = 16000e

- t 10

P = 16000e

- 0 10

P = 16000e

0

P = £16000





b) Calculate the value after 5 years… 5 years implies t = 5 Substitute t = 5 into the formula…

3A

P = 16000e

- t 10

P = 16000e

- t 10

P = 16000e

- 5 10

P = 16000e

-0.5

P = £9704.49





c) What is the implied value of the car in the long run (ie – what value does it tend towards?) Imagine t tends towards infinity (gets really big)

3A

P = 16000e

- t 10

P = 16000e

- t 10

P = 16000 x 0

P = £0

e- t

101

(10√e)t

Bigger t

= Bigger denominator

= Smaller Fraction value…





d) Sketch the Graph of P against t Value starts at £16000 Tends towards 0, but doesn’t get there…

3A

P = 16000e

- t 10

P = 16000e

- t 10

P

t

£16000

t is independent so goes on the x axis

P is dependant on t so goes on the y axis


You need to be able to solve equations

involving natural logarithms and e

This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’

3B

Example Question 1

3xe

ln( ) ln(3)xe

ln( ) ln(3)x e

ln(3)x

1.099x

Take natural logs of both sides

Use the ‘power’ lawln(e) = 1

(e to the power something is e)

Work out the answer or leave as a logarithm

You do not necessarily need to write these

steps…





3B

Example Question 1

2 7xe Take natural logs

Use the power law

2ln( ) ln(7)xe

2 ln(7)x

ln(7) 2x

0.054x

Subtract 2

Work out the answer or leave as a logarithm





3B

Example Question 1

ln 4x ‘Reverse

ln’Work it out if

needed…

4x e

54.598x





3B

Example Question 1

ln(3 2) 3x ‘Reverse

ln’

Add 2

33 2x e 33 2x e

3 2

3

ex

Divide by 3


You need to be able to plot and understand graphs of

the function which is inverse to ex

The inverse of ex is logex (usually written as lnx)

We know from chapter 2 that an inverse function is a

reflection in the line y = x…

3B

y = ex

y = lnx

y = x

(0,1)

(1,0)




3B

y = lnx

(1,0)

y = lnx f(x)

y = 2lnx 2f(x)

y = 2lnx

All output (y) values doubled for the same

input (x) values…




3B

y = lnx

(1,0)

y = lnx f(x)

y = lnx + 2

f(x) + 2

y = lnx + 2

(0.14,0)

ln 2y x

0 ln 2x

2 ln x 2e x

0.13533... x

Let y = 0

Subtract 2

Inverse ln

Work out x!

All output (y) values increased by 2 for the

same input (x) values…




3B

y = lnx

(1,0)y = lnx f(x)

y = -lnx -f(x)y = -lnx

All output (y) values ‘swap sign’ for the

same input (x) values…




3B

y = lnx

(1,0)y = ln(x) f(x)

y = ln(2x) f(2x)

y = ln(2x)

All output (y) values the same, but for half the input (x) values…

(0.5,0)




3B

y = lnx

(1,0)y = ln(x) f(x)

y = ln(x + 2)

f(x + 2)

y = ln(x + 2)

All output (y) values the same, but for

input (x) values 2 less than before

(-1,0)

ln( 2)y x

ln(2)y

0.69314...y

Let x = 0

Work it out (or leave as

ln2)

(0, ln2)




3B

y = lnx

(1,0)y = ln(x) f(x)

y = ln(-x) f(-x)

y = ln(-x)

All output (y) values the same, but for

input (x) values with the opposite sign to

before

(-1,0)




3B

y = lnx

(1,0)


y = 3 + ln(2x)

y = ln(2x)

(0.025,0)

y = 3 + ln(2x)

The graph of ln(2x), moved up 3

spaces…

3 ln(2 )y x 0 3 ln(2 )x 3 ln(2 )x 3 2e x 3

2

ex

Let y = 0

Subtract 3

Reverse ln

Divide by 2




3B

y = lnx

(1,0)


y = ln(3 - x)

The graph of ln(x), moved left 3 spaces, then

reflected in the y axis. You must do the reflection last!

y = ln(3 + x)

(-2,0)

(2,0)

y = ln(3 - x)

ln(3 )y x

ln(3)y Let x = 0

(0,ln3)

x = 3




The number of elephants in a herd can be represented by the equation:

Where n is the number of elephants and t is the time in years after

2003.

a) Calculate the number of elephants in the herd in 2003 Implies t = 0

3B

40150 80t

N e

40150 80t

N e

0

40150 80N e

0150 80N e

150 80N

70N

t = 0

e0 = 1






2003.

b) Calculate the number of elephants in the herd in 2007 Implies t = 4

3B

40150 80t

N e

40150 80t

N e

4

40150 80N e

150 72.4N

77.6 (78)N

t = 4

Round to the nearest whole

number


You need to be able to plot and understand

graphs of the function which is inverse to ex



2003.

c) Calculate the year when the population will first exceed 100

elephants Implies N = 100

3B

40150 80t

N e

40150 80t

N e

40100 150 80t

e

N = 100

4050 80t

e

400.625t

e

ln(0.625)40

t

40ln(0.625) t

18.8 (19 years)t

2003 + 19 = 2022

Subtract 150

Divide by -80

Take natural logs

Multiply by 40


You need to be able to plot and understand

graphs of the function which is inverse to ex



2003.

d) What is the implied maximum number in the herd?

Implies t ∞

3B

40150 80t

N e

40

t

e

40150 80t

N e

40

1t

e

Rearrange

As t increases

Denomintor becomes bigger

Fraction becomes smaller, towards 0

150N




f(x) = x – 1g(x) = ex + 1

a) Calculate gf(4)

3B




f(x) = x – 1g(x) = ex + 1

b) Calculate g-1(x)

3B

Summary• We have learnt a new number, e, and seen what is

stands for and where it is used

• We have plotted the graph of ex and lnx, which are inverse functions

• We have seen how to transform these graphs

• We have solved practical problems involving theese

• We have also seen again how to solve logarithmic equations…

introduction we are going to look at exponential functions we will learn about a new ‘special’...

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