4: the function © christine crisp “teach a level maths” vol. 2: a2 core modules

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4: The function 4: The function x e y © Christine Crisp Teach A Level Maths” Teach A Level Maths” Vol. 2: A2 Core Vol. 2: A2 Core Modules Modules

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Page 1: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

4: The function 4: The function xey

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

Page 2: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

Module C3

Page 3: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay

growth

Functions of this type, with a > 1, are called functions.

xay We’ve already met the functione.g.

xy 2

xy 3xy 4

)1,0(

Autograph demo

Page 4: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay xay We will now investigate the

gradient of Notice first that as x increases, y increases

e.g.

xy 2

),2(41x x

),1(21

x )1,0(x )2,1(

x )4,2(

x )8,3(

Autograph demo

Page 5: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay

dx

dy

dx

dy

xy 2gradient

Notice first that as x increases, y increases. . . and the also increases

e.g.

x x x x

x

x

The gradient function

x x x x x

x

xay We will now investigate the gradient of

looks

It the same asbut . . .

x2

Page 6: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay

xy 2

dx

dy

We will now investigate the gradient of e.g.

x

The gradient function

x 55dx

dy

8ye.g. gradie

nt

Notice first that as x increases, y increases. . . and the also increases

Page 7: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay

xy 2

x

dx

dy)2(69.0

What do you think will happen if we repeat the process for ? xy 3Well, goes up more steeply than so we get a similar result but the gradient function is above the curve.

xy 3 xy 2

Putting the 2 graphs on the same axes . . .

Page 8: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay

It can be shown that

xx

dx

dyy 3)10.1(3

xy 2 xy 3x

dx

dy3)101(

x

dx

dy2)690(

So,

suggesting that there is a value of a between 2 and 3 where the gradient of is equal to .

xay xa

The 1st gradient graph is under the original curve . . .and the 2nd is above the curve . . .

Page 9: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay

gradient of equalsxa

)3(7182 d.p.e

The value of a where the is an irrational number, written as e, where

xay

xay

Page 10: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay

Using a letter for an irrational number isn’t a new idea to you.

xey

gradient of equalsxa

)3(7182 d.p.e

The value of a where the is an irrational number, written as e, where

xay

You have used ( the Greek p ) for )..3(1423 pd

Page 11: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

The Gradient of

xay

xey

gradient of equalsxa

)3(7182 d.p.e

The value of a where the is an irrational number, written as e, where

xay

xx edx

dyey

Page 12: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

More Indices and Logs

xey The function contains the index x, so x is a log.BUT the base of the log is e not 10, so

( since an index is a log )

We write as ( n for natural ) so,

elog ln

yxey x ln

Logs with a base e are called natural logs

We know that yxy x

10log10

yxey ex log

Page 13: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

xey

xy

xey

The Inverse of xey

We can sketch the inverse by reflecting in y = x.

is a one-to-one function so has an inverse function.

xexf )(

Finding the equation of the inverse function is easy!

So, xxf ln)(1

0xN.B. The domain is .

So

xy ln

xy ln

Forwards x e it = f(x)Backwards opposite of e it is ln it

Autograph demo

Page 14: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

SUMMARY xexf )(• is a growth function.

7182e• (3 d.p.)

xey • At every point on , the gradient equals y:

xx edx

dyey

• The inverse of is

xexf )(

xxf ln)(1

( log with base e )is defined for x > 0

onlyxln

xy

xy lnxey

Page 15: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

xey

Can you suggest equations for the unlabelled graphs below?

Both graphs are stretches of .xey

HINT:

Page 16: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

xey

This . . . is a stretch with scale factor 2 parallel to the y-axis.The equation is xey 2

xey 2

Page 17: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

xey

This . . . 21 is a stretch with scale factor

parallel to the x-axis.

The equation is xey 2

xey 2

Page 18: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

Page 19: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Page 20: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

More Indices and Logs

yxey ex log

xey The function contains the index x, so x is a log.BUT the base of the log is e not 10, so

We know that yxy x

10log10 ( since an index is a log )

We write as ( n for natural ) so,

elog ln

yxey x ln

Logs with a base e are called natural logs

Page 21: 4: The function © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

xey

SUMMARY xexf )(• is a growth function.

7182e• (3 d.p.)

xey • At every point on , the gradient equals y:

xx edx

dyey

• The inverse of is

xexf )(

xxf ln)(1

( log with base e )

xey xy ln

xy

is defined for x > 0 only

xln