unit 6: functions - part 2

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MATHEMATICS

Learner’s Study and

Revision Guide for

Grade 12

FUNCTIONS ‐ Part 2

INVERSE FUNCTIONS

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle

Examination Questions by the Department of Basic Education

Preparation for the Mathematics examination brought to you by Kagiso Trust

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Contents

Unit 6

Revision notes 3

The inverse graph 5

Inverse function 10

Examination questions with solution hints and answers 13

More questions from past examination papers 17

Answers 26

How to use this revision and study guide

1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic from your teacher in class or from a textbook. Furthermore, the notes cover all the Mathematics from Grade 10 to Grade 12.

2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.

3. The notes and exercises are followed by questions from past examination papers.

4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.

5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.

6. What follows next are more questions taken from past examination papers.

7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.

8. Finally, don’t be a loner. Work through this guide in a team with your classmates.

Functions – Part 2

REVISION UNIT 6: THE INVERSE GRAPH AND THE INVERSE FUNCTION

Recall the definitions of the domain and range of a relationship we have loosely called a function.

Domain: all the values of x for which y or )(xf can be evaluated are known as the domain and

are denoted by the symbol fD .

Example 1: If ( )5

3−

=x

xf we can evaluate ( )xf for all real values of x except 5=x . At 5=x

, ( )xf is said to be undefined. The domain of ( )xf can thus be expressed as { }5: −ℜ∈xDf .

Illustration:

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As can be seen in the illustration, at 5=x there is an asymptote which in this case is a line the

graph of ( )xf neither crosses nor touches.

Range: all the values of y or )(xf which correspond to the values of x in the fD are known as

the range and are denoted by fR .

In the graphical illustration of the above example, it can be seen that y or ( )xf takes all real

values except 0=y which is also an asymptote. Thus { }0: −ℜ∈yRf .


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Example 2: Consider ( ) 52 += xxf .

Illustration:

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f(x) = 2x+5

It can be seen from the graph that ( )xf is defined for all real values of x so that the domain is

( )∞∞−∈ ;: xDf or ℜ∈xDf :

Notice also from the graph that y or ( )xf takes all real values so that the range is

( )∞∞−∈ ;: yRf or ℜ∈yRf :

Restriction of the domain

Instead of defining ( )xf for all teal values of x from ∞− to ∞ , we can restrict the domain.

Example 3: Consider ( ) 52 += xxf defined over the domain [ )1;4−∈x .

Recall that using a bracket like this means that 14 <≤− x . In other words, x takes all the values between ‐4 and 1 including ‐4 but excluding 1. The interval [ )1;4− is said to be closed at

the left hand side end and open at the opposite end. Graphically, a closed end of the interval is demoted by • and an open end by o .


Illustration:

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f(x) = 2x+5

By inspecting the above graph, it can be seen that the range of ( )xf is given by

73: <≤− yRf or [ )7;3: −∈yRf

THE INVERSE GRAPH

The inverse graph is the reflection of a graph in the line xy = .

Illustration:

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f(x) = 3x-7

y=x

Line of reflection

Mirror

f-1(x)=(x+7)/3


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How to find the equation of the inverse graph

In the illustration above, look at the coordinates of a few points on the line 52 += xy and the

coordinates of their reflections across the mirror line xy = :

Line 52 += xy Reflection in mirror line xy = (0; 5) (‐1; 3) (‐2; 1) (‐3; ‐1)

(5; 0) (3; ‐1) (1; ‐2) (‐1; ‐3)

What do you notice from the above?

( )yx; ( )xy;

As can be seen from the above table, the reflection of a point ( )yx; in the line xy = is given by

the point );( xy . That is, the x and y values are simply interchanged or swapped.

Also clear from the swapping of x and y is that the inverse performs the opposite operation to

the function. The operation takes elements in the domain of the function and makes them elements of the range, and vice a versa.

The swapping of x and y in an equation will give us the equation of the inverse graph. The

steps to be taken in finding the equation of the inverse are shown in the box below:

Finding the equation of the inverse

Step 1: Swap the x and y

Step 2: Make y the subject of the equation Example 4: Apply these steps to finding the inverse of 73 −= xy

Solution: Step 1: Swap the x and y : 73 −= yx Step 2: Make y the subject of the equation:

yx 37 =+

so ( )731

+= xy


Notation for the inverse

The inverse of f or ( )xf is denoted by 1−f or ( )xf 1− .

In the above example:

if ( ) 73 −= xxf or 73: −→ xxf

then ( ) ( )7311

+=−

xxf or ( )731:1 +→− xxf

Graphical illustration:

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f(x) = 3x-7

y=x

Line of reflection

Mirror

f-1(x)=(x+7)/3

Example 5: We have previously considered ( ) 52 += xxf and worked out its inverse to be

( ) ( )5211 −=− xxf . Let us now restrict its domain to [ )1;4: −∈xDf . With the help of a sketch,

answer the following questions:

1. What is the range of ( )xf ?

2. What is the domain of ( )xf 1− ?

3. What is the range of ( )xf 1− ?


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Sketch:

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f

f-1

y=x

1. Range of ( )xf

By inspection of the graph the range is given by [ )7;3: −∈yRf .

2. Domain of ( )xf 1−

By inspection of the inverse graph the domain is [ )7;3:1 −∈− xDf

3. Range of ( )xf 1−

By inspection of the inverse graph the range is [ )1;4:1 −∈− yRf


A summary of the above results is shown in the box below:

52: +→ xxf ( ) 2/5:1 −→− xxf Domain [ )1;4− [ )7;3− Range [ )7;3− [ )1;4−

What can clearly be seen in the box is that the domain of f is the range of 1−f ; and the range

of f is the domain of 1−f . That is , the domain and range make a swap in the same way the xand y values of the coordinates swap on reflection in the line xy = , confirming an earlier observation that the inverse performs an opposite operation to the function. Function: We have tended to speak about functions in a loose manner. But in Mathematics a function has a specific definition. A function is a relationship in which each x value corresponds to only one y value. This is said to be a one‐to‐one relationship.

A function is also defined if more than one x take on the same y value. This is said to be a many‐to‐one relationship.

A relationship between x and y does not define a function if each x has more than one y

value. This is said to be a one‐to‐many relationship.

Vertical line test: A quick and easy way to test whether an expression is a function or not is to draw a graph and then a vertical line to cut its graph. If all vertical lines cut the graph only once then the equation of the graph is a function.

Example 6: Use the vertical line test to find if ( ) 2xxf = is a function.

Solution: The graph of ( ) 2xxf = is shown below.

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A vertical line is drawn anywhere to cut the graph as shown above.


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The vertical line cuts the graph of ( ) 2xxf = at one and only one point.

Therefore ( ) 2xxf = is a function. INVERSE FUNCTION

An inverse function is an inverse that is also a function. Apply the vertical line test to the graph

of 1−f to see if the inverse is indeed a function.

Example 7: Find the inverse of ( ) 2xxf = and find out if the inverse is a function.

Solution: Replace ( )xf by y so that 2xy =

Step 1: Swap the x and y : 2yx =

Step 2: Make y the subject of the equation: xy ±=

Thus the inverse is ( ) xxf ±=−1

But the inverse of 2xy = is not a function because, as the ± double sign indicates, for every

value of x there are two of y . It is a one‐to‐many value relationship.

That xy ±= is not a function can also be seen by drawing its graph and applying the vertical

line test to the graph:

To draw the graph of xy ±= we reflect the graph of 2xy = in the line xy = :

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y=x

As can be seen, the graph of the inverse is cut at two points by any vertical line confirming that the inverse is not a function.


Changing the domain of a function in order to make its inverse a function as well

If we now define ( ) 2xxf = so that its domain is restricted to 0≥x then the inverse ( )xf 1− will

be a function as can be verified from the sketch below:

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y=x


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Example 8: Find the inverse of xy 3=

Step 1: Swap the x and y : yx 3= Step 2: Make y the subject of the equation: xy 3log=

The inverse of xy 3= is a function as can be verified by sketching it and applying the vertical line

test.

Use these axes to sketch the graphs of xy 3= and its inverse. Remember to show the line of reflection. Draw a vertical line to verify that the inverse is indeed a function.

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PAPER 1 QUESTION 5 DoE/ADDITIONAL EXEMPLAR 2008


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PAPER 1 QUESTION 5 DoE/ADDITIONAL EXEMPLAR 2008

Number Hints and answers Work out the solutions in the boxes below 5.1 Given a point on the curve, you

should be able to work out the value of a by substituting the coordinates of the point into the equation of the curve. Work out a . Answer:

21

=a

5.2 What do you have to do to determine the equation of the inverse of a function? Swap x and y , and make y the subject. Answer:

xy21log= or xy 2log−=

or 21log 2=y

5.3 The answer is clear from a rough sketch of the inverse function. So use the space opposite to make a rough sketch. Answer:

80 << x

5.4 Go for it by turning the verbal expression of the transformation into a mathematical expression! Answer:

( )3

21 −

⎟⎠⎞

⎜⎝⎛=

x

xq or ( ) 32 +−= xxq


PAPER 1 QUESTION 5 DoE/NOVEMBER 2008


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PAPER 1 QUESTION 5 DoE/NOVEMBER 2008

Number Work out the solutions in the boxes below 5.1 Sketch the graphs below. DIAGRAM SHEET 1 5.2

5.3

Answer: You write down the answer.

5.4

You are given ( )xh .

So what is ⎟⎠⎞

⎜⎝⎛ +

21xh ? This means replace x in

( )xh by 21

+x .

What is ( )xh2 ? This means multiply ( )xh by 2. You may have to simplify the resulting expressions to show that they are identical. Do this and you will have shown what you are required to show.


MORE QUESTION FROM PAST EXAMINATION PAPERS

EXEMPLAR 2008


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Preparatory Examination 2008


Feb – March 2009


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DIAGRAM SHEET 1


November 2009 (Unused paper)


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DIAGRAM SHEET 1


November 2009 (1)


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Feb – March 2010

DIAGRAM SHEET 2


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ANSWERS

Exemplar 2008

6.1 21

=a

6.2 xy21log=

6.3 Not a function because ( ) 1-or 11 =− xg

6.4 [ )∞∈ ;0x or ( ]0;∞−∈x 6.5.1 10 << x 6.5.2 0=x Preparatory Examination 2008 7.1.1 xy 4log=

7.1.2 1−f is a function because only one correseponding y value for every x value.

7.1.3 ( )x

xh ⎟⎠⎞

⎜⎝⎛=

41

7.2 4=a

7.3 ( ) 12

4−

−=

xxm

7.4 x ‐intercept is (6; 0) y ‐intercept is (0; ‐3)

Feb/March 2009 7.1 ( )1;0Q

7.2 2=a 7.3 xy 2log= 7.4 Sketch:

7.5 5,0>x 7.6 36,11−=x


Feb/March 2009 7.1 ( )1;0Q

7.2 2=a 7.3 xy 2log= 7.4 Sketch:

7.5 5,0>x 7.6 36,11−=x

November 2009 (unused papers) 5.1 2=b 5.2 ( )2;1D is the turning point of g .

5.3 xy 2log= 5.4 Sketch:

5.5 ( ) 2xxh −=

5.6 0≤x or 0≥x 5.7 Maximum value = 4 November 2009(1) 6.1

6.2 ( ) 2+= xxh 6.3 2−= xy 6.4


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November 2009 (1) 8.1 0>x or ( )∞∈ ;0x

8.2 xy −= 2 or x

y ⎟⎠⎞

⎜⎝⎛=

21

8.3 0=y 8.4.1 Reflect the graph of f over the x ‐axis Or For each point the y ‐coordinate changes sign. 8.4.2 Reflect the graph of f over the line xy = . Then shift the graph down 5 units. (This answer can be expressed in other ways as well). 8.5 80 << x or ( )8;0∈x Feb/March 2010 7.1 xy 3log=

7.2 Sketch:

7.3 52 << x

unit 6: functions - part 2

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