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Chapter 6 – Functions 06 Learning Outcomes

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1 Chapter 6 – Functions 06 Learning Outcomes
In this chapter you have learned to: Recognise that a function assigns a unique output to a given input Form composite functions Graph linear, quadratic, cubic, exponential and logarithmic functions Interpret equations of the form f(x) = g(x) as a comparison of the above functions Use graphical methods to find approximate solutions to f(x) = 0, f(x) = k and f(x) = g(x) Express quadratic functions in completed square form Use the completed square form to find roots, turning points and to sketch the function Apply transformations to selected functions Recognise injective, surjective and bijective functions Find the inverse of a bijective function Sketch the inverse of a function given the function's graph

2 06 Functions In a school, Class 5.1 choose ice-skating, 5.2 choose go-karting, 5.3 choose ice-skating and 5.4 choose paint-balling. These choices can be represented by a function, as illustrated in the mapping diagram: 5.1 5.3 5.2 5.4 Cinema Ice-skating Go-karting Paint-balling Bowling 5.1 is an example of an input. Ice-skating is an example of an output. The domain is the set {5.1, 5.2, 5.3, 5.4} The range is the set {ice-skating, go-karting, paint-balling}. The codomain is the set {cinema, ice-skating, go-karting, paint-balling, bowling}. Example of an Algebraic Function 𝑓 π‘₯ = π‘₯ 2 , π‘₯βˆˆβ„€, βˆ’2≀π‘₯β‰₯ Note: 𝑓 π‘₯ = π‘₯ 2 is often written as 𝑓 π‘₯ ⟼ π‘₯ 2 This function can be shown in a number of ways -6 -5 -4 -3 -2 =1 1 2 3 4 5 6 8 10 12 x y Graph 4 1 -2 -1 2 3 9 Domain/Range Diagram Set of Couples {(-2,4), (-1,1), (0,0), (1,1), (2,4), (3,9)}

3 06 Functions Linear Functions
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y 𝒇 𝒙 =πŸπ’™+𝟏 π’ˆ 𝒙 =βˆ’πŸπ’™+πŸ‘ The graph of a linear function is a straight line. If the coefficient of the variable. is positive the slope of the line will be positive. If the coefficient of the variable. is negative the slope of the line will be negative. The constant value in a linear function dictates where the graph cuts the y-axis. Quadratic Functions -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y π’ˆ 𝒙 = 𝒙 𝟐 βˆ’πŸ’ If the coefficient of the squared variable. is positive the graph will be βˆͺ-shaped. If the coefficient of the squared variable. is negative the graph will be ∩-shaped. π’ˆ 𝒙 =βˆ’ 𝒙 𝟐 +𝟏 A quadratic function may cross the x-axis at a maximum of two points.

4 Transformations of Linear and Quadratic Functions
06 Functions Transformations of Linear and Quadratic Functions -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y 𝒇 𝒙 =πŸπ’™+πŸ’ 𝒇 𝒙 =πŸπ’™+𝟏 𝒇 𝒙 =πŸπ’™βˆ’πŸ π’ˆ 𝒙 = 𝒙 𝟐 βˆ’πŸ’ π’ˆ 𝒙 = (𝒙+πŸ‘) 𝟐 βˆ’πŸ’ π’ˆ 𝒙 = (π’™βˆ’πŸ) 𝟐 βˆ’πŸ’ π’ˆ 𝒙 = 𝒙 𝟐 +𝟐 π’ˆ 𝒙 = 𝒙 𝟐 βˆ’πŸ’ -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y π’ˆ 𝒙 = 𝒙 𝟐 βˆ’πŸ” π’ˆ 𝒙 = 𝒙 𝟐 βˆ’πŸ’ π’ˆ 𝒙 = (𝒙+πŸ‘) 𝟐 βˆ’πŸ‘ π’ˆ 𝒙 = (π’™βˆ’πŸ‘) 𝟐 +𝟏

5 06 Functions Expressing Quadratic Functions in Completed Square Form (Vertex Form) To write f(x) = x2 – 4x + 7 in completed square form, we need to β€˜complete the square’. To complete the square, half the x coefficient which is –2. The square of –2 is 4. Now add and subtract 4 to the function to give f(x) = x2 – 4x + 4 βˆ’ 4 + 7 This simplifies to give f(x) = x2 – 4x -3 -2 -1 1 2 3 4 5 6 7 8 x y π’ˆ 𝒙 = 𝒙 𝟐 βˆ’πŸ’π’™+πŸ• f(x) = (x2 – 4x + 4) + 3 f(x) = (x βˆ’ 2)2 + 3 The co-ordinates of this vertex is (2,3). 𝑽𝒆𝒓𝒕𝒆𝒙 (𝟐,πŸ‘) To summarise: If a > 0 β‡’ βˆͺ -shaped Vertex is a minimum point If a < 0 β‡’ ∩-shaped Vertex is a maximum point y = a(x – h)2 + k Axis of symmetry x = h Vertex (h,k)

6 06 Functions Cubic Functions
𝒇 𝒙 = 𝒙 πŸ‘ + 𝒙 𝟐 βˆ’πŸ’π’™βˆ’πŸ’ -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y 𝒇 𝒙 = βˆ’π’™ πŸ‘ βˆ’πŸ‘ 𝒙 𝟐 +𝒙+πŸ‘ If the coefficient of x3 is positive, the graph will start low and end high. If the coefficient of x3 is negative, the graph will start high and end low. A cubic function may cross the x-axis at a maximum of three points. Exponential Functions -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y 𝒇 𝒙 = 𝟐 𝒙 π’ˆ 𝒙 = 𝟐 βˆ’π’™ The graph of the function f(x) = abx will pass through the point (0,a). The graph of an exponential function will never touch or cross the x-axis.

7 06 Functions Logarithmic Functions
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y Exponential functions have an inverse function. We call these inverses: logarithmic functions. The logarithmic function is the image of the exponential function under axial symmetry in the line y = x π’š=𝒙 𝒇 𝒙 = 𝟐 𝒙 π’ˆ 𝒙 = π’π’π’ˆ 𝟐 𝒙 Composite Functions f(x) = 6x + 2 and g(x) = x3, where both f and g are functions that map from R to R. Find the value of f o g(x) where f o g(x) is a composite function. π‘“π‘œπ‘” π‘₯ =𝑓(𝑔(π‘₯)) 𝑓 π‘₯ =6π‘₯+2 To find 𝑓 𝑔 π‘₯ replace every x in the 𝑓 function with 𝑔(π‘₯) 𝑓 𝑔 π‘₯ =6(𝑔(π‘₯))+2 𝑓 𝑔 π‘₯ =6( (π‘₯ 3 ))+2 π‘“π‘œπ‘” π‘₯ =6 π‘₯ 3 +2

8 Injective, Surjective and Bijective Functions
06 Functions Injective, Surjective and Bijective Functions f is an injective function, as every output has a unique input. It is a 1 to 1 relationship. g is a surjective (β€˜onto’) function, as every element in B is an output. p is a bijective function, as it is both injective and surjective (β€˜1 to 1’ and β€˜onto’). Horizontal Line Test for Injectivity Horizontal Line Test for Surjectivity -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 y x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y 𝒇 𝒙 =𝒙+πŸ‘ π’ˆ 𝒙 = 𝒙 𝟐 βˆ’πŸ’ Any horizontal line drawn will never cut the graph of f at more than one point. This shows that the function f is injective. Any horizontal line drawn will cut the graph of g on at least one point. This shows that the function g is surjective.

9 06 Functions Inverse Functions
Definitions If two functions f and g are defined so that 𝑓:Aβ†’B and 𝑔:Bβ†’A, then, if π‘“π‘œπ‘” π‘₯ = π‘”π‘œπ‘“ (π‘₯)=π‘₯, we say that 𝑓 and 𝑔 are inverse functions of each other. NOTE: A function 𝑓 has an inverse function 𝑓 βˆ’1 if and only if 𝑓 is a bijection. NOTE: The domain and codomain are important in determining if a function is invertible or not. Example: If we define a function β„Ž such that β„Ž:Rβ†’R:π‘₯β†’ 𝑒 π‘₯ then β„Ž is not bijective (it is injective but not surjective) and so it has no inverse. However, if we define h to be β„Ž:𝑅→ 𝑅 + :π‘₯β†’ 𝑒 π‘₯ then β„Ž is bijective (invertible), with β„Ž βˆ’1 defined as β„Ž βˆ’1 : 𝑅 + β†’ 𝑅 : π‘₯β†’lnπ‘₯. Investigate if the function 𝒇:𝐑→𝐑:𝒙→ 𝒙 πŸ‘ +𝟏 has an inverse. -4 -3 -2 -1 1 2 3 4 x y The horizontal line test shows that the function is injective. The function is also surjective, as every element in the codomain has a corresponding domain value. ∴ The function is a bijection and therefore has an inverse.


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