Higher Functions Unit 1.2

A rule which changes one number into another. What is a FUNCTION A rule which changes one number into another. A rule which links each member of Set A to EXACTLY ONE member of Set B.

Domain and Range The values that make up the set of independent values are the domain The values that make up the set of dependent values are the range. Make up and example of a relation then define the domain and range of the relation

Consider the function f(x) = 2x + 1 The values that make up the set of independent values are the domain Let x = -2 , 0 , 5 (the domain) The values that make up the set of dependent values are the range. Make up and example of a relation then define the domain and range of the relation f(x) = -3 , 1 , 11 (the range)

Consider the function f(x) = 2x + 1 Set A Set B –2 5 –3 1 11 Make up and example of a relation then define the domain and range of the relation Domain Range

Some Things About Functions 1. Remember that for each member of the DOMAIN there is EXACTLY one value in the RANGE. 2. Sometimes you have to restrict the numbers in the DOMAIN depending on the function you are working with.

Pick some values for x, the independent values Domain and Range To draw a graph of y = 3x + 1 Pick some values for x, the independent values Use the values for x in y = 3x + 1 to find the corresponding values of y, the dependent values. Make up and example of a relation then define the domain and range of the relation The y values depend on the values chosen for x which are independent

Functions - Notation There are two ways to write a function but they both have the same effect The function on the left is read as follows:- ‘Function f such that x is mapped onto x2 + 4’ The function on the right is read as follows:- ‘f of x equals x2 + 4’

Suppose that f(x) = 3x - 2 and g(x) = x2 +1 Example 1 f(2) = 3×2 – 2 = 4 Suppose that f(x) = 3x - 2 and g(x) = x2 +1 g(4) = 42 + 1 =17 g(2) = 22 + 1 = 5 f(5) = 5×3-2 = 13 f(1) = 3×1 - 2 =1 f(1) = 3×1 - 2 =1 (a) g( f(2) ) = g(4) = 17 g(26) = 262 + 1 = 677 g(5) = 52 + 1 = 26 (b) f( g (2) ) = f(5) = 13 (c) f( f(1) ) = f(1) = 1 (d) g( g(5) ) = g(26) = 677

Suppose that f(x) = x + 5 and g(x) = x2 – 3 Example 1 Suppose that f(x) = x + 5 and g(x) = x2 – 3 (a) g( f(1) ) = g(6) = 33 (b) f( g (1) ) = f(– 2) = 3 (c) f( f(3) ) = f(8) = 13 (d) g( g(2) ) = g(1) = – 2

Suppose that f(x) = x + 3 and g(x) = 2x – 3 Example 2 finding a formula Suppose that f(x) = x + 3 and g(x) = 2x – 3 (a) g( f(x) ) = g(x + 3 ) = 2(x + 3) – 3 = 2x + 6 – 3 = 2x + 3 (b) f( g(x) ) = f(2x – 3) = (2x – 3) + 3 = 2x – 3 + 3 = 2x

We need to revise some algebra before going any further There is a quicker way to square brackets than using FOIL (x + a)2 = (x + a)(x + a) = x2 + ax + ax + a2 1. Square each term = x2 + 2ax + a2 x2 + a2 2. Multiply the two terms then double your answer x2 + 2ax + a2

2. Multiply the two terms then double your answer (x + a)2 (x + 5)2 1. Square each term x2 + a2 x2 + 25 2. Multiply the two terms then double your answer x2 + 2ax + a2 x2 + 10x + 25 2 × 5 × x

2. Multiply the two terms then double your answer (3x – 2)2 1. Square each term x2 + a2 9x2 + 4 2. Multiply the two terms then double your answer x2 + 2ax + a2 9x2 – 12x + 4 2 × (–2) × 3x

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)) = g(3x – 2) = (3x – 2)2 + 1 = 9x2 – 12x + 5 (b) f(g(x)) = f(x2 + 1) = 3(x2 + 1) – 2 = 3x2 + 1 NB: g(f(x))  f(g(x)) in general. CHECK g(f(2)) = 9 × 22 - 12 × 2 + 5 = 36 - 24 + 5 = 17 f(g(2)) = 3 × 22 + 1 = 13

Page 24 4, 5, 6, 7 up to 7d

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)) Put x2 – 6x + 13 = 8 = g(x – 3) then x2 – 6x + 5 = 0 = (x – 3)2 + 4 or (x – 5)(x – 1) = 0 = x2 – 6x + 13 So x = 1 or x = 5 CHECK g(h(5)) = g(2) = 22 + 4 = 8

Choosing a suitable Domain f(x) = 2x + 1 Set A Set B –2 5 –3 1 11 Make up and example of a relation then define the domain and range of the relation Domain Range

Choosing a Suitable Domain There are times when a function f(x) does not exist for specific values of x. In particular, when the function involves algebraic fractions or square roots. A fraction does not exist when the denominator is zero A square root does not exist for values less than zero

Choosing a Suitable Domain (i) Suppose f(x) = 1 x2 – 4 f(x) is not defined when x2 – 4 is zero x2 – 4 = 0  is the set of Real Numbers So x2 = 4 So x = –2 or 2 Hence domain is {x   : x  –2 or 2 } f(x) exists for all values of x except when x = –2 or 2

Choosing a Suitable Domain Suppose f(x) = 1 x2 – 1 and g(x) = 2x function is not defined when (2x)2 – 1 is zero  (2x + 1)(2x – 1) = 0  x = ± ½ f(g(x)) exists for all x except x = ± ½

f(x) exists only if 2x – 1 is greater than or equal to zero Choosing a Suitable Domain f(x) exists only if 2x – 1 is greater than or equal to zero 2x – 1 ≥ 0 2x ≥ 1 x ≥ 1/2 f(x) exists only if x ≥ 1/2

Functions with Fractions To simplify expressions like this change all parts to the same denominator x ≠ 1

f(f(x)) exists for all x except x = – 2 Suppose f(x) = 1 x + 1 Find an expression for f(f(x)) and a suitable domain. f(f(x)) exists for all x except x = – 2

Always give suitable Domain Page 20 ~ 8 to 12 Page 24 ~ 7f) to 9 Always give suitable Domain

Many examples require you to sketch a parabola to find where values are positive or negative

Greater than or equal to zero when x ≤ -4 and x ≥ 2 (ii) Suppose that g(x) = (x2 + 2x - 8) We need (x2 + 2x – 8)  0 -4 2 Suppose x2 + 2x – 8 = 0 Then (x + 4)(x – 2) = 0 So x = – 4 or x = 2 A “happy” parabola So domain = { x   : x  – 4 or x  2 }

x ≠ 0 x ≠ – 3

Find f(g(x)) and g(f(x)) in their simplest form when Exercise Find f(g(x)) and g(f(x)) in their simplest form when Find f(g(x)) and g(f(x)) in their simplest form when

Inverse Functions Reversing the process

Consider the function f(x) = 2x + 1 Set A Set B –2 5 –3 1 11 The function which changes the numbers in Set B back into the numbers in set A is called an INVERSE function Make up and example of a relation then define the domain and range of the relation The opposite of 2x + 1 is take 1 and divide by 2

Consider the function f(x) = 2x + 1 Set A Set B –2 5 –3 1 11 Make up and example of a relation then define the domain and range of the relation g(x) is the inverse of f(x), usually written as f -1(x)

g(x) is the inverse of f(x), usually written as f -1(x) You have seen this notation before when using your calculator sin-1, cos-1, tan-1 are the inverse functions of sin, cos and tan Make up and example of a relation then define the domain and range of the relation

Graphically the inverse function is the reflection of the given function in the line y = x [ f(x) = 2x + 1] Make up and example of a relation then define the domain and range of the relation

Exponential & Log Functions

Exponential (to the power of) Graphs A function in the form f(x) = ax where a > 0, a ≠ 1 is called an exponential function to base a . Consider f(x) = 2x x -3 -2 -1 0 1 2 3 f(x) 1/8 ¼ ½ 1 2 4 8 Remember a0 = 1 and a1 = a

Exponential (to the power of) Graphs y = 2x A growth function (1 , 6) (2 , 4) y = 6x (1 , 2)

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

To obtain y from x we must ask the question Log Graphs ie x 1/8 ¼ ½ 1 2 4 8 y -3 -2 -1 0 1 2 3 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 “the logarithm to base 2 of x” y = log2x or “log to the base 2 of x”

y = log2x (2 , 1) NB: x > 0 Major Points (i) y = log2x passes through the points (1,0) & (2,1) . As x  , y   but at a very slow rate and as x  0 y  – ∞

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

The following questions are on Graphs & Functons Non-calculator questions will be indicated

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. a) b)

a) Find expressions for: i) f(g(x)) ii) g(f(x)) Functions f and g are defined on suitable domains by f(x)= sin x and g(x) = 2x. a) Find expressions for: i) f(g(x)) ii) g(f(x)) b) Solve 2 f(g(x)) = g(f(x)) for 0 ≤ x ≤ 360 a) b)

Functions are defined on a suitable set of real numbers. a) Find expressions for b) i) Show that ii) Find a similar expression for and hence solve the equation a) b) Now use exact values equation reduces to

a) b) c) from the graph, k  4 A sketch of the graph of y = f(x). The graph has a maximum at A and a minimum at B(3, 0) a) Find the co-ordinates of the turning point at A. b) Hence, sketch the graph of Indicate the co-ordinates of the turning points. c) Write down the range of values of k for which g(x) = k has 3 real roots. a) Differentiate for SP, f(x) = 0 when x = 1 t.p. at A is: b) Graph is moved 2 units to the left, and 4 units up t.p.’s are: c) For 3 real roots, line y = k has to cut graph at 3 points from the graph, k  4

a) Find b) If find in its simplest form. a) b)

a) b) c) Functions 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. c) Find the area enclosed between this graph and the x-axis. a) b) c) Graph cuts x axis at 0 and 1 Now evaluate Area

The functions f and g are defined on a suitable domain by a) Find an expression for b) Factorise a) b) Difference of 2 squares Simplify