KT Classroom Unit 6: Functions - Part 2

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 1

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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

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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.

2

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 f (x ) can be evaluated are known as the domain and are denoted by the symbol D f .

3 we can evaluate f ( x ) for all real values of x except x = 5 . At x = 5 x−5 , f ( x ) is said to be undefined. The domain of f ( x ) can thus be expressed as D f : x ∈ ℜ − {5} .

Example 1: If f (x ) =

Illustration:

y 9 8 7 6 5 4 3 2 1

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As can be seen in the illustration, at x = 5 there is an asymptote which in this case is a line the graph of f ( x ) neither crosses nor touches.

Range: all the values of y or f (x ) which correspond to the values of x in the D f are known as the range and are denoted by R f .

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

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

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Example 2: Consider f ( x ) = 2 x + 5 .

Illustration: y 9 8

f(x) = 2x+5 7 6 5 4 3 2 1

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

D f : x ∈ (− ∞; ∞ )

Df : x ∈ ℜ

or

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

R f : y ∈ (− ∞; ∞ )

Rf : y ∈ ℜ

or

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

Example 3: Consider f ( x ) = 2 x + 5 defined over the domain x ∈ [− 4;1) .

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

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Functions – Part 2

Illustration:

y 9 8 7 6 5

f(x) = 2x+5 4 3 2 1

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

R f : −3 ≤ y < 7

R f : y ∈ [− 3;7 )

or

THE INVERSE GRAPH The inverse graph is the reflection of a graph in the line y = x .

Illustration: y 9 8 7 6 5 4 3 2

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

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Line of reflection Mirror

f(x) = 3x-7

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y=x

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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 y = 2 x + 5 and the coordinates of their reflections across the mirror line y = x : Line y = 2 x + 5 (0; 5) (‐1; 3) (‐2; 1) (‐3; ‐1)

Reflection in mirror line y = x (5; 0) (3; ‐1) (1; ‐2) (‐1; ‐3)

What do you notice from the above?

(x; y ) ( y; x )

As can be seen from the above table, the reflection of a point ( x; y ) in the line y = x is given by the point ( y; x ) . 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 y = 3 x − 7 Solution:

Step 1: Swap the x and y : x = 3 y − 7 Step 2: Make y the subject of the equation:

x + 7 = 3y so

y=

1 (x + 7 ) 3 6

Functions – Part 2

Notation for the inverse The inverse of f or f ( x ) is denoted by f

−1

or f

−1

(x ) .

In the above example:

if f (x ) = 3x − 7 or f : x → 3 x − 7

then f

−1

(x ) = 1 (x + 7 ) or f −1 : x → 1 (x + 7 ) 3

3

Graphical illustration:

y 9 8 7 6 5 4 3 2

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

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Line of reflection Mirror y=x

f(x) = 3x-7

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Example 5: We have previously considered f ( x ) = 2 x + 5 and worked out its inverse to be 1 (x − 5) . Let us now restrict its domain to D f : x ∈ [− 4;1) . With the help of a sketch, 2 answer the following questions: f −1 ( x ) =

1. What is the range of f ( x ) ?

2. What is the domain of f −1 ( x ) ?

3. What is the range of f −1 (x ) ?

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Sketch:

y 9 8 7 6

y=x

5 4

f

3 2 1

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1. Range of f ( x ) By inspection of the graph the range is given by R f : y ∈ [− 3;7 ) .

2. Domain of f −1 (x ) By inspection of the inverse graph the domain is D f −1 : x ∈ [− 3;7 ) 3. Range of f −1 ( x ) By inspection of the inverse graph the range is R f −1 : y ∈ [− 4;1)

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Functions – Part 2

A summary of the above results is shown in the box below: f −1 : x → ( x − 5) / 2 [− 3;7) [− 4;1)

f : x → 2x + 5

[− 4;1) [− 3;7)

Domain Range

What can clearly be seen in the box is that the domain of f is the range of f −1 ; and the range of f is the domain of f −1 . That is , the domain and range make a swap in the same way the x and y values of the coordinates swap on reflection in the line y = x , 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 f ( x ) = x 2 is a function.

Solution: The graph of f (x ) = x 2 is shown below. y 9 8 7 6 5 4 3 2 1

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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 f ( x ) = x 2 at one and only one point.

Therefore f ( x ) = x 2 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 f

−1

to see if the inverse is indeed a function.

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

Solution:

Replace f ( x ) by y so that y = x 2 Step 1: Swap the x and y : x = y 2 y=± x

Step 2: Make y the subject of the equation:

Thus the inverse is

f −1 ( x ) = ± x

But the inverse of y = x 2 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 y = ± x 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 y = ± x we reflect the graph of y = x 2 in the line y = x : y 9 8 7 6

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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.

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Functions – Part 2

Changing the domain of a function in order to make its inverse a function as well If we now define f ( x ) = x 2 so that its domain is restricted to x ≥ 0 then the inverse f −1 (x ) will be a function as can be verified from the sketch below:

y 9 8 7 6

f y=x

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Example 8: Find the inverse of y = 3 x Step 1: Swap the x and y : x = 3 y y = log 3 x Step 2: Make y the subject of the equation:

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

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

y 9 8 7 6 5 4 3 2 1

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Functions – Part 2

PAPER 1 QUESTION 5

DoE/ADDITIONAL EXEMPLAR 2008

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PAPER 1 QUESTION 5 Number 5.1

Hints and answers 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:

a= 5.2

Work out the solutions in the boxes below

1

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: y = log 1 x or y = − log 2 x

2

or y = log 2 5.3

5.4

1

2

The answer is clear from a rough sketch of the inverse function.

So use the space opposite to make a rough sketch.

Answer: 0 < x < 8 Go for it by turning the verbal expression of the transformation into a mathematical expression!

Answer:

⎛1⎞ ⎝2⎠

q(x ) = ⎜ ⎟

x −3

or q ( x ) = 2 − x +3

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DoE/ADDITIONAL EXEMPLAR 2008

Functions – Part 2

PAPER 1 QUESTION 5

DoE/NOVEMBER 2008

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PAPER 1 QUESTION 5

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

5.3

5.4

Answer: You write down the answer.

You are given h( x ) .

1⎞ ⎛ ⎟ ? This means replace x in 2⎠ ⎝ 1 h( x ) by x + .

2

So what is h⎜ x +

What is 2h( x ) ? This means multiply h( x ) 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.

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DoE/NOVEMBER 2008

Functions – Part 2

MORE QUESTION FROM PAST EXAMINATION PAPERS EXEMPLAR 2008

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Preparatory Examination 2008

18

Functions – Part 2

Feb – March 2009

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DIAGRAM SHEET 1

20

Functions – Part 2

November 2009 (Unused paper)

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DIAGRAM SHEET 1

22

Functions – Part 2

November 2009 (1)

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24

Functions – Part 2

Feb – March 2010

DIAGRAM SHEET 2

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ANSWERS Exemplar 2008 1 6.1 a = 2 6.2 y = log 1 x 2

6.3 Not a function because g −1 ( x ) = 1 or - 1 6.4 x ∈ [0; ∞ ) or x ∈ (− ∞;0]

6.5.1 0 < x < 1 6.5.2 x = 0

Preparatory Examination 2008 7.1.1 y = log 4 x 7.1.2 f −1 is a function because only one correseponding y value for every x value.

Feb/March 2009 7.1 Q(0;1) 7.2 a = 2 7.3 y = log 2 x 7.4 Sketch:

x

⎛1⎞ 7.1.3 h( x ) = ⎜ ⎟ ⎝4⎠

7.2 a = 4 4 7.3 m( x ) = −1 x−2 7.4 x ‐intercept is (6; 0) y ‐intercept is (0; ‐3)

7.5 x > 0,5 7.6 x = −11,36

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Functions – Part 2

5.5 h( x ) = − x 2 5.6 x ≤ 0 or x ≥ 0

Feb/March 2009 7.1 Q(0;1) 7.2 a = 2 7.3 y = log 2 x 7.4 Sketch:

5.7 Maximum value = 4

November 2009(1) 6.1

6.2 h( x ) = x + 2 6.3 y = x − 2

6.4

7.5 x > 0,5 7.6 x = −11,36

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

5.3 y = log 2 x

5.4 Sketch:

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

x

⎛1⎞ 8.2 y = 2 or y = ⎜ ⎟ ⎝2⎠ 8.3 y = 0 −x

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 y = x .

Then shift the graph down 5 units.

(This answer can be expressed in other ways as well).

8.5 0 < x < 8 or x ∈ (0;8)

Feb/March 2010 7.1 y = log 3 x

7.2 Sketch:

7.3 2 < x < 5

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