Unit 2 EF 1.3 Higher Higher Maths Composite Functions Exponential and Log Graphs Graph...

Unit 2 EF 1.3

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Higher

Higher MathsHigher Maths

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Composite Functions

Exponential and Log Graphs

Graph Transformations

Trig Graphs

Inverse functionMindmap

Exam Question Type

Derivative Graphs f’(x)

Completing the Square

Solving equations / /Inequations

Unit 2 EF 1.3

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Higher

Graph Transformations

We will investigate f(x) graphs of the form

1. f(x) ± k

2. f(x ± k)

3. -f(x)

4. f(-x)

5. kf(x)

6. f(kx)

Each affect the

Graph of f(x) in a certain

0 2 4 6 8x

-2-4-6

Transformation f(x) ± k

(x , y) (x , y ± k)

Mapping

f(x) + 5 f(x) - 3

Transformation f(x) ± k

Keypoints

y = f(x) ± k

moves original f(x) graph vertically up or down

+ k move up

- k move down

Only y-coordinate changes

NOTE: Always state any coordinates given on f(x)

on f(x) ± k graph

f(x) - 2

A(-1,-2) B(1,-2)C(0,-3)

f(x) + 1

B(90o,0)

A(45o,0.5)

C(135o,-0.5)

B(90o,1)

A(45o,1.5)

C(135o,0.5)

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3C

0 2 4 6 8x

-2-4-6

Transformation f(x ± k)

(x, y) (x ± k , y)

Mapping

f(x - 2) f(x + 4)

Transformation f(x ± k)

Keypoints

y = f(x ± k)

moves original f(x) graph horizontally left or right

+ k move left

- k move right

Only x-coordinate changes

on f(x ± k) graph

0 2 4 6 8x

-2-4-6

Transformation -f(x)

(x, y) (x , -y)

Mapping

Flip inx-axis

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3E

Transformation -f(x)

Keypoints

y = -f(x)

Flips original f(x) graph in the x-axis

y-coordinate changes sign

on -f(x) graph

- f(x)

A(-1,0) B(1,0)

C(0,1)

- f(x)

B(90o,0)

A(45o,0.5)

C(135o,-0.5)A(45o,-0.5)

C(135o,0.5)

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3G

0 2 4 6 8x

-2-4-6

Transformation f(-x)

(x, y) (-x , y)

Mapping

Flip iny-axis

Transformation f(-x)

Keypoints

y = f(-x)

Flips original f(x) graph in the y-axis

x-coordinate changes sign

on f(-x) graph

B(0,0)

C’(-1,1)

A’(1,-1)A(-1,-1)

C (1,1)

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3I

0 2 4 6 8x

-2-4-6

Transformation kf(x)

(x, y) (x , ky)

Mapping

Stretch iny-axis

2f(x) 0.5f(x)

Compress iny-axis

Transformation kf(x)

Keypoints

y = kf(x)

Stretch / Compress original f(x) graph in the

y-axis direction

y-coordinate changes by a factor of k

on kf(x) graph

0 2 4 6 8x

-2-4-6

Transformation f(kx)

(x, y) (1/kx , y)

Mapping

Compress inx-axis

f(2x) f(0.5x)

Stretch inx-axis

Transformation f(kx)

Keypoints

y = f(kx)

Stretch / Compress original f(x) graph in the

x-axis direction

x-coordinate changes by a factor of 1/k

on f(kx) graph

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3K & 3M

Unit 2 EF 1.3

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Higher

You need to be able to work with combinations

Combining Transformations

(-1,-3)

2f(x) + 1

f(0.5x) - 1

f(-x) + 1

-f(x + 1) - 3

Explain the effect the following have

(a)-f(x)

(b)f(-x)

(c)f(x) ± k

(d)f(x ± k)

(e)kf(x)

(f)f(kx)

Name :

(-1,-3)

f(x + 1) + 2

-f(x) - 2

(-1,-3)

(1,-2)

(1,3)(-1,4)

(-1,-3)(1,-5)

(-1,1)

(-1,-3)

2f(x) + 1

f(0.5x) - 1

f(-x) + 1

-f(x + 1) - 3

(a)-f(x) flip in x-axis

(b)f(-x) flip in y-axis

(c)f(x) ± k move up or down

(d)f(x ± k) move left or right

(e)kf(x) stretch / compressin y

direction

(e)f(kx) stretch / compress

in x direction

Name :

(-1,-3)

(-2,-1)

f(x + 1) + 2

-f(x) - 2

(-2,0)

(0,-6)

(-1,-3)

(-1,-5)

(-2,-4)(-1,-3)

The diagram shows the graph of a function f.

f has a minimum turning point at (0, -3) and a

point of inflexion at (-4, 2).

a) sketch the graph of y = f(-x).

b) On the same diagram, sketch the graph of y = 2f(-x)

Graphs & Functions Higher

a) Reflect across the y axis

b) Now scale by 2 in the y direction-1 3 4

y = f(-x)

y = 2f(-x)

Part of the graph of is shown in the diagram.

On separate diagrams sketch the graph of

Indicate on each graph the images of O, A, B, C, and D.

( )y f x

( 1)y f x 2 ( )y f x

graph moves to the left 1 unit

graph is reflected in the x axis

graph is then scaled 2 units in the y direction

(2, 1)

(2, -1)

(2, 1)

y=f(x)

y= -f(x)

y= 10 - f(x)

Graphs & Functions Higher =

a) On the same diagram sketch

i) the graph of

ii) the graph of

b) Find the range of values of x for

which is positive

2( ) 4 5f x x x

( )y f x

10 ( )y f x

10 ( )f x

2( 2) 1x

b) Solve:210 ( 2) 1 0x

2( 2) 9x ( 2) 3x 1 or 5x

10 - f(x) is positive for -1 < x < 5

A sketch of the graph of y = f(x) where is shown.

The graph has a maximum at A (1,4) and a minimum at B(3, 0)

Sketch the graph of

Indicate the co-ordinates of the turning points. There is no need to

calculate the co-ordinates of the points of intersection with the axes.

3 2( ) 6 9f x x x x

( ) ( 2) 4g x f x

Graph is moved 2 units to the left, and 4 units up(3, 0) (1, 4)

(1, 4) ( 1, 8)

t.p.’s are:

(-1,8)

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Outcome 3Higher

Trig Graphs

The same transformation rules

apply to the basic trig graphs.

NB: If f(x) =sinx then 3f(x) = 3sinx

and f(5x) = sin5x

Think about sin replacing f !

Also if g(x) = cosx then g(x) – 4 = cosx – 4

and g(x + 90) = cos(x + 90) Think about cos replacing g !

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Outcome 3Higher

Sketch the graph of y = sinx - 2

If sinx = f(x) then sinx - 2 = f(x) - 2

So move the sinx graph 2 units down.

y = sinx - 2

Trig Graphs

090o 180o 270o 360o

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Outcome 3Higher

Sketch the graph of y = cos(x - 50)

If cosx = f(x) then cos(x - 50) = f(x - 50)So move the cosx graph 50 units right.

Trig Graphs

y = cos(x - 50)o

90o 180o 270o 360o

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Outcome 3Higher

Trig Graphs

Sketch the graph of y = 3sinx

If sinx = f(x) then 3sinx = 3f(x)

So stretch the sinx graph 3 times vertically.

y = 3sinx

90o 180o 270o 360o

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Outcome 3Higher

Trig Graphs

Sketch the graph of y = cos4x

If cosx = f(x) then cos4x = f(4x)

So squash the cosx graph to 1/4 size horizontally

y = cos4x

090o 180o 270o 360o

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Outcome 3Higher

Trig Graphs

Sketch the graph of y = 2sin3xIf sinx = f(x) then 2sin3x = 2f(3x)So squash the sinx graph to 1/3 size horizontally and also double its height.

y = 2sin3x

360o180o 270o

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Trig Graph Trig Graph

090o 180o 270o 360o

Write down equations for

graphs shown ?

CombinationsHigher

y = 0.5sin2xo + 0.5

y = 2sin4xo- 1

Write down the equations in the form f(x) for the graphs shown?

y = 0.5f(2x) + 0.5

y = 2f(4x) - 1

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Trig GraphsTrig Graphs

090o 180o 270o 360o

Combinationsy = cos2xo + 1

y = -2cos2xo - 1Higher

Write down the equations for the graphs shown?

Write down the equations in the form f(x) for the graphs shown?

y = f(2x) + 1y = -2f(2x) - 1

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 4A & 4B

Show-me boards

Unit 2 EF 1.3

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Higher

A function in the form f(x) = ax where a > 0, a ≠ 1

is called an exponential function to base a .

Exponential (to the power of) Graphs

Exponential Functions

Consider f(x) = 2x

x -3 -2 -1 0 1 2 3

f(x) 1 1/8 ¼ ½ 1 2 4 8

Unit 2 EF 1.3

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Higher

The graph of

y = 2x

(0,1)(1,2)

Major Points

(i) y = 2x passes through the points (0,1) & (1,2)

(ii) As x ∞ y ∞ however as x -∞ y 0 .(iii) The graph shows a GROWTH function.

Unit 2 EF 1.3

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Higher

y -3 -2 -1 0 1 2 3

x 1/8 ¼ ½ 1 2 4 8

To obtain y from x we must ask the question

“What power of 2 gives us…?”

This is not practical to write in a formula so we say

y = log2x“the logarithm to base 2 of x”

or “log base 2 of x”

Log Graphs

Unit 2 EF 1.3

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Higher

The graph of

y = log2x (1,0)

Major Points

(i) y = log2x passes through the points (1,0) & (2,1) .(ii) As x ∞ y ∞ but at a very slow rate

and as x 0 y -∞ .

NB: x > 0

Unit 2 EF 1.3

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Higher

The graph of y = ax always passes through (0,1) & (1,a)

It looks like ..

y = ax

Exponential (to the power of) Graphs

Unit 2 EF 1.3

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Higher

The graph of y = logax always passes through (1,0) & (a,1)

It looks like ..

y = logax

Log Graphs

Unit 2 EF 1.3

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Higher

f-1(x) = logax

Connection

f(x) = ax

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 2H

HHM Ex 3N , 3O and 15K

HHM Ex 3P

0 2 4 6 8x

-2-4-6

Derivative f’(x)

All to do with

GRADIENT !

f’(x)

Unit 2 EF 1.3

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Higher

This is a method for changing the format

of a quadratic equation

so we can easily sketch or read off key information

Completing the square format looks like

f(x) = a(x + b)2 + c

Warning ! The a,b and c values are different

from the a ,b and c in the general quadratic function

Unit 2 EF 1.3

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Higher

Half the x term and square the

coefficient.

Complete the square for x2 + 2x + 3

and hence sketch function.

f(x) = a(x + b)2 + c

x2 + 2x + 3

(x2 + 2x + 1) + 3 Compensate

(x + 1)2 + 2

-1Tidy up !

Unit 2 EF 1.3

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Higher

sketch function.f(x) = a(x + b)2 + c

= (x + 1)2 + 2

Mini. Pt. ( -1, 2) (-1,2)

Unit 2 EF 1.3

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Higher

2(x2 - 4x) + 9 Half the x term and square the

coefficient.

Take out coefficient of

x2 term.Compensate !

Complete the square for 2x2 - 8x + 9

f(x) = a(x + b)2 + c

2x2 - 8x + 9

2(x2 – 4x + 4) + 9 Tidy up

2(x - 2)2 + 1

Unit 2 EF 1.3

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Higher

= 2(x - 2)2 + 1

Mini. Pt. ( 2, 1)

Unit 2 EF 1.3

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Higher

Half the x term and square the

coefficient

Take out coefficient of x2

compensate

Complete the square for 7 + 6x – x2

f(x) = a(x + b)2 + c

-x2 + 6x + 7

-(x2 – 6x + 9) + 7 Tidy up

-(x - 3)2 + 16

a = -1

c = 16

+ 9-(x2 - 6x) + 7

Unit 2 EF 1.3

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Higher

= -(x - 3)2 + 16

Mini. Pt. ( 3, 16)

(3,16)

Given , express in the form

Hence sketch function.

Quadratic Theory Higher

2( ) 2 8f x x x ( )f x 2x a b

2( ) ( 1) 9f x x

(-1,9)

(0,-8)

Quadratic Theory Higher

a) Write in the form

b) Hence or otherwise sketch the graph of

2( ) 6 11f x x x 2x a b

( )y f x

a) 2( ) ( 3) 2f x x

b) For the graph of 2y x moved 3 places to left and 2 units up.

minimum t.p. at (-3, 2) y-intercept at (0, 11)

(-3,2)

(0,11)

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 8D

19 Apr 202319 Apr 2023 Created by Mr. Lafferty@www.mathsrevision.comCreated by Mr. Lafferty@www.mathsrevision.com

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Solving Quadratic Equations

Nat 5 Examples

Solve ( find the roots ) for the following

x(x – 2) = 0

x = 0 andx - 2 = 0

4t(3t + 15) = 0

4t = 0 and 3t + 15 = 0

t = -5 t = 0 and

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Nat 5 Examples

x2 – 4x = 0

x(x – 4) = 0

x = 0 andx - 4 = 0

16t – 6t2 = 0

2t(8 – 3t) = 0

2t = 0 and 8 – 3t = 0

t = 8/3 t = 0 and

Common

Factor

Common

Factor

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Nat 5 Examples

x2 – 9 = 0

(x – 3)(x + 3) = 0

x = 3 and x = -3

100s2 – 25 = 0

25(2s – 1)(2s + 1) = 0

2s – 1 = 0 and2s + 1 = 0

s = - 0.5s = 0.5 and

Difference 2 squares

Take out common

factor

25(4s2 - 1) = 0

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Nat 5 Examples

2x2 – 8 = 0

2(x2 – 4) = 0

x = 2 and x = - 2

80 – 125e2 = 0

5(16 – 25e2) = 0

4 – 5e = 0 and4 + 5t = 0

e = - 4/5 e = 4/5 and

Common

Factor

Common

Factor

2(x – 2)(x + 2) = 0(x – 2)(x + 2) = 0

5(4 – 5e)(4 + 5e) = 0(4 – 5e)(4 + 5e) = 0

(x – 2) = 0 and(x + 2) = 0

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Nat 5 Examples

x2 + 3x + 2 = 0

(x + 2)(x + 1) = 0

x = - 2 andx = - 1

SAC Method

x + 2 = 0 x + 1 = 0 and

3x2 – 11x - 4 = 0

(3x + 1)(x - 4) = 0

x = - 1/3 and x = 4

SAC Method

3x + 1 = 0 andx - 4 = 0

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Nat 5 Examples

x2 + 5x + 4 = 0

(x + 4)(x + 1) = 0

x = - 4 andx = - 1

SAC Method

x + 4 = 0 x + 1 = 0 and

1 + x - 6x2 = 0

(1 + 3x)(1 – 2x) = 0

x = - 1/3 and x = 0.5

SAC Method

1 + 3x = 0 and1 - 2x = 0

Unit 2 EF 1.3

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Higher

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When we cannot factorise or solve graphically quadratic equations we need to use the

quadratic formula.

2-b ± (b -4ac)x =

ax2 + bx + c

Quadratic FormulaQuadratic Formula

Unit 2 EF 1.3

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Higher

Example : Solve x2 + 3x – 3 = 0

2-(3) ± ((3) -4(1)(-3))x =

ax2 + bx + c

1 3 -3

-3 ± 21=

Unit 2 EF 1.3

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Higher

-3 + 21x =

2-3 - 21

x = 0.8 x = -3.8

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Nat 5 Examples

√ both sides

√4 = ± 2

(x – 3)2 – 4 = 0

(x – 3)2 = 4

x – 3 = ± 2

x = 3 ± 2

x = 5 x = 1and

√ both sides √of 7 = ±

(x + 2)2 – 7 = 0

(x + 2)2 = 7

x + 2 = ± √7

x = -2 ± √7

x = -2 + √7 x = -2 - √7and

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 8E and Ex8G

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Higher

Solving Quadratic Inequations

To solve inequations ( inequalities) the steps are

1. Solve the equation = 0

2. Sketch graph

3. Read off solution

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Higher

Solve the inequation x2 + 5x – 6 > 0

1. Solve the equation = 0 2. Sketch graph

x2 + 5x – 6 = 0

(x - 1)(x + 6) = 0

x = 1 and x = - 6

x < -6 and x > 1

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 8F and Ex8K

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Higher

Solve the inequation 3 - 2x – x2 < 0

1. Solve the equation = 0 2. Sketch graph

3 - 2x – x2 = 0

(x + 3)(x - 1) = 0

x = - 3 and x = 1

x < -3 and x > 1

x2 + 2x – 3 = 0

Unit 2 EF 1.3

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Higher

If a function f(x) has roots/zeros at a, b and c

then it has factors (x – a), (x – b) and (x – c)

And can be written as f(x) = k(x – a)(x – b)(x – c).

Functions from Graphs

Unit 2 EF 1.3

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Higher

Example

-2 1 5

y = f(x)

Finding a Polynomial From Its Zeros

Unit 2 EF 1.3

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Higher

f(x) has zeros at x = -2, x = 1 and x = 5,

so it has factors (x +2), (x – 1) and (x – 5)

so f(x) = k (x +2)(x – 1)(x – 5)

f(x) also passes through (0,30) so replacing x by 0

and f(x) by 30 the equation becomes

30 = k X 2 X (-1) X (-5)

ie 10k = 30

ie k = 3

Unit 2 EF 1.3

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Higher

Formula is f(x) = 3(x + 2)(x – 1)(x – 5)

f(x) = (3x + 6)(x2 – 6x + 5)

f(x) = 3x3 – 12x2 – 21x + 30

Quad Demo Cubic Demo

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 7H

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What are Functions ?

Functions describe how one quantity

relates to another

Car Part

Assembly line

Defn: A function or mapping is a relationship between two sets in which each member of the first set is connected to exactly one member in the

second set.

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What are Functions ?

Functions describe how one quantity

relates to another

Washing Machine

OutputInputyx

Functionf(x)

y = f(x)

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Defining a Functions

A function can be thought of as the relationship between

Set A (INPUT - the x-coordinate)

SET B the y-coordinate (Output) .

Unit 2 EF 1.3

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Higher

YFunction !

Functions & GraphsFunctions & Graphs

Unit 2 EF 1.3

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Higher

Function & Graphs Function & Graphs

YFunction !

Unit 2 EF 1.3

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Higher

YNot a

function !!

Cuts graph

more than once !

Function & GraphsFunction & Graphs

x must map to

one value of y

Unit 2 EF 1.3

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Higher

Functions & GraphsFunctions & Graphs

Y Not a function !!

Cuts graph

more than once!

Unit 2 EF 1.3

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Higher

Functions & MappingsFunctions & Mappings

A function can be though of as a black box

x - Coordinate

Domain

Members (x - axis)Co-Domain

Members (y - axis)

Function

Output

y - Coordinatef(x) = x2+ 3x - 1

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Finding the Function

Find the output or input values for the functions below :

f(x) = x2

7f(x) = 4x - 1

f(x) = 3x

Examples

Unit 2 EF 1.3

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Higher

Functions & MappingFunctions & Mapping

Functions can be illustrated in three ways:

1) by a formula.

2) by arrow diagram.

3) by a graph (ie co-ordinate diagram).

ExampleSuppose that f: A B is defined by

f(x) = x2 + 3x where A = { -3, -2, -1, 0, 1}.FORMULA

then f(-3) = 0 , f(-2) = -2 , f(-1) = -2 , f(0) = 0 ,f(1) = 4

NB: B = {-2, 0, 4} = the range!

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The standard way to represent a function

is by a formula.

Function Notation

Examplef(x) = x + 4

We read this as “f of x equals x + 4”

“the function of x is x + 4

f(1) = 5 is the value of f at 1

f(a) = a + 4 is the value of f at a

1 + 4 = 5

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For the function h(x) = 10 – x2.

Calculate h(1) , h(-3) and h(5)

h(1) =

Examples

h(-3) = h(5) =

h(x) = 10 – x2

Function Notation

10 – 12 = 9

10 – (-3)2 =

10 – 9 = 1

10 – 52 =

10 – 25 = -15

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For the function g(x) = x2 + x

Calculate g(0) , g(3) and g(2a)

g(0) =

Examples

g(3) = g(2a) =

g(x) = x2 + x

Function Notation

02 + 0 =

32 + 3 =

(2a)2 +2a =

4a2 + 2a

Unit 2 EF 1.3

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Higher

COMPOSITION OF FUNCTIONS

( or functions of functions )

Suppose that f and g are functions where

f:A B and g:B C

with f(x) = y and g(y) = z

where x A, y B and z C.

Suppose that h is a third function where

h:A C with h(x) = z .

Composite FunctionsComposite Functions

Unit 2 EF 1.3

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Higher

x y zf g

We can say that h(x) = g(f(x))

“function of a function”

Unit 2 EF 1.3

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Higher

Composite FunctionsComposite Functionsf(2)=3x2 – 2 =4

g(4)=42 + 1 =17

f(5)=5x3-2 =13Example 1

Suppose that f(x) = 3x - 2 and g(x) = x2 +1

(a) g( f(2) ) = g(4) = 17

(b) f( g (2) ) = f(5) = 13

(c) f( f(1) ) = f(1) = 1

(d) g( g(5) ) = g(26)= 677

f(1)=3x1 - 2 =1

g(26)=262

+ 1 =677

g(2)=22 + 1 =5

f(1)=3x1 - 2 =1

g(5)=52 + 1 =26

Unit 2 EF 1.3

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Higher

Suppose that f(x) = 3x - 2 and g(x) = x2 +1

Find formulae for (a) g(f(x)) (b) f(g(x)).

(a) g(f(x)) = ( )2 + 1

= 9x2 - 12x + 5

(b) f(g(x)) = 3( ) - 2= 3x2 + 1

g(f(2)) = 9 x 22 - 12 x 2 + 5

= 36 - 24 + 5= 17

f(g(2)) = 3 x 22 + 1= 13

NB: g(f(x)) f(g(x)) in general.

3x - 2 x2 +1

Unit 2 EF 1.3

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Higher

Let h(x) = x - 3 , g(x) = x2 + 4 and k(x) = g(h(x)). If k(x) = 8 then find the value(s) of x.

k(x) = g(h(x))

= ( )2 + 4

= x2 - 6x + 13

Put x2 - 6x + 13 = 8

then x2 - 6x + 5 = 0

or (x - 5)(x - 1) = 0

So x = 1 or x = 5

Unit 2 EF 1.3

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Higher

Choosing a Suitable Domain

(i) Suppose f(x) = 1 . x2 - 4

Clearly x2 - 4 0

So x2 4

So x -2 or 2

Hence domain = {xR: x -2 or 2 }

Unit 2 EF 1.3

srevis

Higher

(ii) Suppose that g(x) = (x2 + 2x - 8)

We need (x2 + 2x - 8) 0

Suppose (x2 + 2x - 8) = 0

Then (x + 4)(x - 2) = 0

So x = -4 or x = 2

So domain = { xR: x -4 or x 2 }

Composite FunctionsComposite FunctionsSketch graph

The functions f and g are defined on a suitable domain by

a) Find an expression for b) Factorise

2 2( ) 1 and ( ) 2f x x g x x

( ( ))f g x ( ( ))f g x

a) 22 2( ( )) ( 2) 2 1f g x f x x

2 22 1 2 1x x Difference of 2 squares

Simplify 2 23 1x x

Functions and are defined on suitable domains.

a) Find an expression for h(x) where h(x) = f(g(x)).

b) Write down any restrictions on the domain of h.

( ) 2 3g x x

( ( )) (2 3)f g x f x a)1

2 3 4x

( )2 1

b) 2 1 0x 1

3( ) 3 ( ) , 0x

f x x and g x x

a) Find

b) If find in its simplest form.

( ) where ( ) ( ( ))p x p x f g x

3( ) , 3

xq x x

( ( ))p q x

3( ) ( ( ))x

p x f g x f

x 3 3x

3( 1)x

( ( )) x

p q x p

33 3x x

9 3(3 ) 3

Graphs & Functions HigherFunctions f and g are defined on the set of real numbers by

a) Find formulae for

i) ii)

b) The function h is defined by

Show that and sketch the graph of h.

2( ) 1 and ( )f x x g x x

( ( ))f g x ( ( ))g f x

( ) ( ( )) ( ( ))h x f g x g f x

2( ) 2 2h x x x

2 2( ( )) ( ) 1f g x f x x 2( ( )) ( 1) 1g f x g x x

22( ) 1 1h x x x 2 2( ) 1 2 1h x x x x 22 2x x

Unit 2 EF 1.3

srevis

Higher

Inverse FunctionsInverse Functions

A Inverse function is simply a function in reverse

Function

Outputf(x) = x2+ 3x - 1

InputOutputf-1(x) = ?

Unit 2 EF 1.3

srevis

Higher

Inverse Function

Find the inverse function given

f(x) = 3x

Example

Remember

f(x) is simply the

y-coordinate

y = 3x

Using Changing the subject

rearrange into

Rewrite replacing y with

This is the inverse function

f-1(x) =x

Unit 2 EF 1.3

srevis

Higher

Inverse Function

f(x) = x2

Example

Remember

f(x) is simply the

y-coordinate

y = x2

rearrange into

x = √y

f-1(x) = √x

Unit 2 EF 1.3

srevis

Higher

Inverse Function

f(x) = 4x - 1

Example

Remember

f(x) is simply the

y-coordinate

y = 4x - 1

rearrange into

f-1(x) =

Unit 2 EF 1.3

srevis

Higher

Are you on Target !

• Update you log book

• Make sure you complete and correct MOST of the Composite Functionquestions in the past paper booklet.

Graphs & Functions

y = -f(x)

y = f(-x)

y = f(x) ± k

y = f(kx)

Move verticallyup or downs

depending on k

flip iny-axis

flip inx-axis

- Stretch or compressvertically

depending on k

y = kf(x)

Stretch or compress

horizontally depending on k

f(x)f(x)

f(x)y = f(x ± k)

Move horizontallyleft or right

depending on k

Remember we can combine

these together !!

0 < k < 1 stretch

k > 1 compress

0 < k < 1 compress

k > 1 stretch

Composite Functions

A complex function made up of 2 or

more simpler functions

f(x) = x2 - 4 g(x) = 1x

Domainx-axis valuesInput

Rangey-axis valuesOutput

x2 - 41

x2 - 4

Restriction x2 - 4 ≠ 0

(x – 2)(x + 2) ≠ 0

x ≠ 2 x ≠ -2

g(f(x)) g(f(x)) =

f(x) = x2 - 4g(x) = 1x

Domainx-axis valuesInput

Rangey-axis valuesOutput

f(g(x))

Restriction x2 ≠ 0

2- 4 =

Similar to composite

Write down g(x) with brackets for x

g(x) =1

inside bracket put f(x)

g(f(x)) =1

x2 - 4

- 41x2

f(g(x)) =Write down f(x) with brackets for x

f(x) = ( )2 - 4

inside bracket put g(x)

f(g(x)) =1

Functions & Graphs

TYPE questions(Sometimes Quadratics)

SketchingGraphs

CompositeFunctions

Steps :

1.Outside function staysthe same EXCEPT replacex terms with a ( )

2. Put inner function in bracket

You need to learn basic movements

Exam questionsnormally involve two movements

Remember orderBODMAS

Restrictions :

1.Denominator NOT ALLOWEDto be zero

2.CANNOT take the square rootof a negative number

Unit 2 EF 1.3 Higher Higher Maths Composite Functions Exponential and Log Graphs Graph...

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