C3: Functions. Learning objectives


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1 CHAPTER C3: Functions Learning objectives After studing this chapter ou should: be familiar with the terms oneone and manone mappings understand the terms domain and range for a mapping understand the term function be able to find the range of a function be able to form composite functions understand the condition for an inverse function to eist.. Notation In the first chapter of C ou were introduced to function notation. It is rather like having a machine into which numbers are fed, and for each value input, the machine determines the output value. The function f which squares the number input and then adds 3 to the result can be represented b f() 2 3 so that f() 3 4 and f( 4) The epression f() is sometimes called the image of. The letter f is frequentl used to represent a function, since it is the first letter of the word function, but it is quite in order to use an other letter instead. However, when ou have two different functions, it is usual to call the first one f and the second one g, and so on. Input function machine utput An alternative notation for f, often used in universit tets, is f: 2 3. However this notation will not be used in the AQA eaminations. Worked eample. The functions f and g are defined for all real values of and are such that f() 2 4 and g() 4. (a) Find f( 3) and g(0.3). (b) Find the two values of for which f() g().
2 2 C3: Functions Solution (a) f( 3) ( 3) g(0.3) (4 0.3) (b) Since f() g() ou can write Therefore Factorising gives ( 5)( ) 0, so that 5 or. Hence the two values of for which f() g() are 5 and. EXERCISE A The function f is defined for all real values of b f() 2 3. Find the values of: (a) f( ), (b) f(3). 2 The function g is defined for all real values of b g() (2) 3. Find the values of: (a) g( ), (b) g(3). 3 Given that s() 3 2sin, find: (a) s(0 ), (b) s(90 ), (c) s(30 ), (d) s(270 ). 4 Given that t() 4 tan, find: (a) t(0 ), (b) t(45 ), (c) t(80 ), (d) t(35 ). 5 The functions f and g are defined for all real values of and are such that f() 3 5 and g() 4. (a) Find f( ) and g(2). (b) Find the value of for which f() g(). 6 The functions f and g are defined for all real values of and are such that f() 2 2 and g() 5 2. (a) Find f( 3) and g( 5). (b) Find the two values of for which f() g(). 7 Given that f() , evaluate: (a) f(0), (b) f(), (c) f(2), (d) f( ), (e) f( 2). 8 Given that g() ( 3) 3, evaluate: (a) g(0), (b) g(), (c) g(2), (d) g( ), (e) g( 2), (f) g( 4), (g) g( 2.99), (h) g(b), (i) g(a 3). 9 Given that f() (2 5), find the eact values of: (a) f(0), (b) f(), (c) f(2), (d) f( ), (e) f( 2).
3 C3: Functions 3.2 Mapping diagram Instead of finding a single value of f(), imagine that each number in the set { 2, 0,, 3, 4} is input in turn to a function machine. The corresponding output values could be represented as a mapping diagram as shown in the diagram. Can ou recognise what the mapping is actuall doing? Each number is cubed and then one is added to the result. It doesn t matter that no elements are mapped onto 74 and 3. Because each element of the first set is mapped to eactl one element of the second set, we sa the mapping is oneone. The set of input values is called the domain. So for this mapping the domain is the set { 2, 0,, 3, 4}. If ou map from the domain using arrows, the set of values where the arrows map onto is called the range. Here, the range is the set { 7,, 2, 28, 65}. Consider a second mapping diagram as shown. The larger set on the right that contains the range is called the codomain so that, in this eample, the codomain is the set { 3, 7,, 2, 28, 65, 74}. The term codomain will not be used in eamination questions Notice that both of the numbers 3 and 5 are mapped onto 2. Also the numbers 0 and 2 are mapped onto 3. This time the domain is the set { 5, 3, 2, 0, 3} and the range is the set { 3, 0, 2}. In this case more than one element of the domain maps onto the same element in the range. The mapping is manone. When a mapping is oneone or manone it is called a function. It is usuall represented b a single letter such as f, g, or h, etc. The set of numbers for which a function is defined is called the domain. A mapping such as illustrated below is oneman and cannot represent a function
4 4 C3: Functions A function f consists of two things: a defining rule such as f() 2 3; its domain. The set of values the function can take for a given domain is called the range. Worked eample.2 The function h has domain { 2,, 0, 3, 7} and is defined b h() ( 3) 2 2. Find the range of h. Solution h( 2) ( 2 3) h( ) ( 3) h(0) (0 3) h(3) (3 3) h(7) (7 3) The range of h is {2,, 8, 27} Because h( ) and h(7) give the same value, we onl write the value 8 once in the range..3 Functions with continuous intervals as domains It is more common for a function to have an interval of values as its domain rather than the domain consisting of just a finite set of values such as { 2, 0,, 3}. Suppose the function f is defined for the domain 2 3 b f() 3 2. The graph of 3 2 is a straight line. The section of the line ou are restricted to is where 2 3, since this is the domain of the function f. Since f( 2) and f(3) 9 2, the onl part of the line to be considered is the section between the two points with coordinates ( 2, 4) and (3, ). (3, ) The graph of f() is drawn opposite. 4 ( 2, 4) It is a good idea to put blobs at the end points to remind ou that these are values the function can actuall take.
5 C3: Functions 5 The possible values that can take are therefore 4. This defines the range of the function. The range of f can be written as 4 or 4 f(). When the domain of f is a continuous interval, the range can be found b considering the graph of f(). The range consists of the possible values that can take. The range of f is written as an inequalit involving f(). Worked eample.3 The function g has domain 2 and is defined b g() 2. (a) Sketch the graph of g(). (b) Find the range of g. Solution (a) The quadratic graph 2 has a minimum point at (0, ). It is useful to evaluate the function at the end points of the domain. Here g( ) 0 and g(2) 4 3. The section of the parabola required is sketched below and the end points are indicated b small blobs. 3 (2, 3) The graph of g() is shown opposite. (, 0) (b) You need to consider more than the two end points when considering the range of values that can take. Notice that the graph comes down as low as. The greatest value can take is 3, as given b the righthand etremit of the graph. Hence the range of g is given b g() 3. EXERCISE B The function f has domain { 2,, 0,, 2, 3} and is defined b f() 2 2. Find the range of f. 2 The function g has domain { 2,, 0,, 2} and is defined b g() (2 ) 3 7. Find the range of g.
6 6 C3: Functions 3 The function h is defined for 3 b h() 3. Find the range of h. 4 The function f is defined for 2 b f() 2. (a) Sketch the graph of f(). (b) Find the range of f. 5 The function g is defined for 4 b g() 0 2. (a) Sketch the graph of g(). (b) Find the range of g. 6 The function f is defined for all real values of b f() ( 2)( 2). (a) (i) Find the coordinates of the points where the graph of f() cuts the coordinate aes. (ii) Sketch the graph of f(). (b) State the range of f. [A].4 Further eamples involving domain and range Sometimes the domain is defined in such a wa as to eclude the end points of the interval. For instance, the domain ma be of the form 3, or perhaps something like 2. Worked eample.4 The function g with domain 2 is defined b g() 3 4. Sketch the graph of g() and find the range of g. Solution 4 (2, 4) You can use a graphics calculator or recognise that the basic graph of 3 has been translated through 0. 4 (, 5) 5 You need to find the smallest and greatest values of g() from the graph. g( ) ( ) g(2) (even though 2 is not in the domain) It is important to sketch the graph onl for the domain indicated. There is a difference between the graph of 3 4, which eists for all real values of, and the graph of g() which is sketched here.
7 C3: Functions 7 Since the function is not defined when 2 it is a good idea to represent this in some special wa. This is usuall done b drawing a small circle to remind ou that the point (2, 4) is not actuall included in the graph. The graph shows that can take all values between 5 and 4. The function can take the value 5 but not the value 4. Hence, the range of g is given b 5 g() 4. Worked eample.5 The function f is defined b f() 4 2, graph of f() and find the range of f. 2. Sketch the Solution It is likel that our first attempt to obtain a sketch will produce something like the one opposite, particularl if ou are using a graphics calculator. However, ou need to restrict the set of values so that onl the part of the graph for > 2 is drawn. The graph of f() for 2 is shown below. Domains ma be defined in this wa in the C3 eamination. Notice that the word domain is not actuall mentioned (2, 0) Although the value 2 is not actuall in the domain ou need to find f(2) As gets larger than 2 ou can see that f() decreases continuousl from the value 0. The range is therefore given b f() 0. EXERCISE C The function g with domain 4 is defined b g() 7. Sketch the graph of g() and find the range of g. You ma wish to use a graphics calculator to help ou with our sketches. 2 The function f with domain 3 is defined b f() 2. Sketch the graph of f() and find the range of f. 3 The function q with domain 2 is defined b q() 3. Sketch the graph of q() and find the range of q.
8 8 C3: Functions 4 The function g is defined b g() 3 5,. Sketch the graph of g() and find the range of g. 5 The function f with domain 2 is defined b f() 3 5. Sketch the graph of f() and find the range of f. 6 The function h with domain 2 3 is defined b h() 2 3. Sketch the graph of h() and find the range of h. 7 The function f with domain is defined b f() 3. Sketch the graph of f() and find the range of f. 8 The function g with domain 3 is defined b g() 6. Sketch the graph of g() and find the range of g. 9 The function f with domain 0 80 is defined b f() 2 sin. Sketch the graph of f() and find the range of f. 0 The function h with domain 0 is defined b h() cos 3. Sketch the graph of h() and find the range of h..5 Greatest possible domain A function is sometimes defined for all real values of. We sa the domain is the set of real numbers,, or. When this is the case, the domain is sometimes omitted and implicitl understood to be the set of real numbers. For instance f() (with no mention of a domain) implies that can take all real values. Sometimes restrictions on the domain are necessar. For instance f() 3 cannot be defined for 0. belongs to. If ou tr to find f(0) on our calculator ou will get an error message. Its greatest possible domain is therefore, 0. Since we cannot find square roots of negative quantities, the function g, where g() 3, does not eist for 3. The greatest possible domain for g is 3. EXERCISE D Determine the greatest possible domain for each of the following functions, f. f() 2 f() 3 ( )
9 C3: Functions 9 3 f() 3 4 f() 2 ( ) f() 6 f() 2 2 ( 4) 3 7 f() 8 f() ( )( 2) 4.6 Graphs that represent functions Consider the two graphs opposite. (a) In graph (a), for each value of ou can draw a vertical line and see that it gives a unique value. An horizontal line for a particular value of also corresponds to a unique value of. Graph (a) represents a oneone function. Repeating the procedure for graph (b). An vertical line gives one value of and so the graph represents a function. This time, however, some of the horizontal lines pass through more than one point on the curve and indicate that more than one value of maps onto a particular value of. (b) The function represented b graph (b) is manone..7 Graphs that do not represent functions Contrast the graphs shown here with those in the previous section. For certain values of ou can draw a vertical line and see that it does not correspond to a unique value of. These graphs do not represent functions. For certain values of there is more than one value of. The corresponding mapping diagram would be oneman and cannot represent the function.
10 0 C3: Functions EXERCISE E For each of the following graphs, state whether it represents a function or not. For the functions, identif them as oneone or manone Composite functions The term composition is used when one operation is performed after another operation. For instance: This function can be written as h() 5( 3). Sometimes ou have two given functions such as f and g and need to perform one function after another. Suppose f() 2 and g() 2 3,. What is f[g()]? g() 2 3 f[g()] f(2 3) (2 3) 2. The epression f[g()] is usuall written without the etra brackets as fg() and fg is said to be a composite function. 3 5( 3) Add 3 Multipl b 5 g() The function gf can be found in a similar wa. f() 2 so that gf() g[f()] = g( 2 ) 2 3 ( 2 ) The composite function gf is such that gf() g f fg() You could tr with a number g(2) f[g(2)] f(8) Although we write fg() the function g operates first on because it is closest to.
11 C3: Functions Worked eample.6 The functions f and g are defined for all real values of b f() 3 5 and g() 3 2. The composite functions fg and gf are such that p = fg and q = gf. Find p() and q(). Solution g() 3 2 f[g()] f[3 2] (3 2) 3 5 The composite function fg p. Hence p() (3 2) 3 5. f() 3 5 g(f()) g( 3 5) 3 2( 3 5) Since gf q, q() 3 2( 3 5) Note that fg is not the same as gf Although we could multipl out the brackets, it is best to leave the functions in this more compact form. EXERCISE F Assume that the domain of the functions in this eercise is the set of real numbers. Find an epression for fg() for each of these functions: (a) f() and g() 5 2, (b) f() 2 3 and g() 2, (c) f() 3 and g() 3, (d) f() 4 2 and g() ( ) 2. 2 For each pair of functions in question, find an epression for gf(). 3 Given that f() 2 2 and g() 3, find: (a) fg(), (b) gf(), simplifing our answers. 4 Given that f() 2 3, find (a) ff(2), (b) ff(a). Solve the equation ff(a) a. 5 Given that f() k 2 and g() 4 3, find in terms of k and : (a) fg(), (b) gf(). State the value of k for which fg() gf(). 6 Given that f() 2 and g() 5, find fg() and gf(). Show that there is a single value of for which fg() gf() and find this value of.
12 2 C3: Functions 7 The functions f and g are defined with their respective domains b 3 f(), 2 2 g() 2. (a) Find the range of g. (b) The domain of the composite function fg is. Find fg() and state the range of fg..9 Domains of composite functions In the eamples of composite functions considered so far the domains have been the real numbers, but sometimes the domains need more careful consideration. Consider the composite function fg g g() The whole of the range of g must be included in the domain of f, otherwise the domain of g needs restricting. f fg() Worked eample.7 Given the functions f and g such that f(), 0, and g() 5,, find the maimum possible domain of fg. Solution It is necessar to solve the inequalit g() 0 so that the range of g consists onl of positive numbers The maimum domain of fg is therefore 5. Initiall it might seem that an real number can be part of the domain of fg since the domain of g is. However g(7) 2, for eample, and this is not acceptable to be fed into f (see the diagram above)..0 Inverse functions The function f defined for all real values of b f() 3 4 can be thought of as a sequence of operations Multipl b 3 Subtract 4 If ou reverse the operations and the flow, Divide b 3 Add 4
13 C3: Functions 3 The new function, g sa, can be written as g() 4. 3 In general fg() f Also gf() g(3 4) (3 4) A function g such that fg() and gf() is said to be the inverse function of f and is denoted b f. In this case, f () 4. 3 Notice that f(3) 5 and g(5) 3. Similarl f( ) 7 and g( 7), etc. This is purel a smbol and should not be thought of as a reciprocal. A reverse flow diagram can be used to find an inverse function when occurs onl once in f(). You consider how f() has been constructed as a sequence of simple operations and set up a flow diagram. Then ou reverse each operation and reverse the direction of the flow to find f (). Worked eample.8 Find: (a) f () and (b) g (), where f() 3 5, and g(),, 2. 2 Solution (a) A flow diagram approach gives which when revised produces 3 5 So f () = 3. 5 Alternativel, f() 3 5 so let 3 5 Rearrange to make the subject of the equation Now interchange and 3 5 Hence f () Cube Add 5 5 Cube root Subtract The reverse flow diagram method can be used to find inverse functions in simple cases when occurs onl once in f(). This worked eample gives a more general method for finding inverse functions. The forward flow diagram has the boes cube then add 5. The reverse flow diagram would be subtract 5 then take the cube root giving the same answer for the inverse function.
14 4 C3: Functions (b) Since occurs more than once in the epression for g(), a flow diagram cannot be used. g() so let 2 2 ( 2) 2 2 ( ) 2 2 ( ) Now interchange and 2 ( ) Hence 2 g () ( ) You can check that the answer is correct b choosing a value of from the domain. For instance, when 3, g() gives g(3) = 3 = 4. Using the answer for g (), g (4) = 8 = = 3, which suggests the answer is correct. Multipl up b ( 2). Collect all the terms involving onto one side. Make the subject of the formula. Essentiall we are interchanging the domain and range to produce an inverse function. The domain for g is,. The inverse of f can be found b the following procedure: Write f(). Rearrange the equation to make the new subject. Interchange and (equivalent to reflecting in ). The new epression for is equal to f ().. Condition for an inverse function to eist In order for f to eist, the function f must be oneone. You can easil draw the graph of f when it eists. The graph of f is obtained from the graph of f() b reflection in the line provided ou have equal scales on the  and aes. This is because the domain and range are interchanged when ou perform an inverse mapping. If a function were manone, the inverse mapping would be oneman and, as ou have seen in Section.7, this could not be a function.
15 C3: Functions 5 Worked eample.9 The function f is defined b f() ( 2) 3, 0 and is sketched with equal scales on the aes. ( 2) 3 (a) Find the range of f. (b) State wh the inverse function f eists. 2 (c) Find f () and sketch the graph of f. (d) State the domain and range of f. Solution (a) f(0) 8 and the graph shows that the values of f() increase as increases. Range is f() 8. (b) For each value of f(), there is a unique value of. The function f is oneone. (c) f() ( 2) 3 so let ( 2) 3 Rearranging to make the new subject. 3 ( 2) 2 3 Interchanging and. 2 3 f () 2 3 Graph of f obtained b reflection of graph of f in line. (d) The domain of f is 8 (since the range of f is f() 8). The range of f is f () 0 (since the domain of f is 0)..2 Selfinverse functions Suppose ou take the reciprocal of 5. You get 0.2. Taking the 5 reciprocal of 0.2 gives 5. Doing the operation twice brings 0.2 ou back to the number ou started with. This is true for ever nonzero number ou tr to take the reciprocal of. Taking the reciprocal is an eample of a selfinverse operation. The function f defined for all nonzero values of b f() is called a selfinverse function. Consider the function g given b g() 2,. If the inverse of g is g, then g g () and g g(). But gg() g(2 ) 2 (2 ) 2 2. This proves that the inverse of g is itself g. Hence g is a selfinverse function. Alternativel g() 2 so let Interchange and. 2 g () 2,
16 6 C3: Functions ther eamples of selfinverse operations are divide into 6 and subtract from 7. Hence, when f() 6, ou can write f () 6 also. Similarl, when g() 7, then g () = 7 as well. You need to remember this if ou are using the flow diagram method to find the inverse function when some operations are selfinverse. Worked eample.0 The function f is defined for all nonzero values of b f() 3 2. Use a reverse flow diagram to find f (). Solution The function f can be thought of as a sequence of operations. If ou reverse the operations and the flow, 2 3 Divide into 2 Subtract from 3 3 Divide into 2 Subtract from 3 The inverse function f is given b f 2 () and is defined 3 for all real values of not equal to 3. EXERCISE G Each of the following functions, f, has domain. Find f () b means of a reverse flow diagram. (a) f() 5 7 (b) f() ( 2) 3 (c) f() (2 ) (d) f() (2 ) 3 6 (e) f() ( 2) 3 (f) f() For each of the functions in question, sketch the graphs of f() and f (). 3 The function f has domain 5 and is defined b 3 f(). 4 (a) Sketch the graph of f(). (b) Find the range of f. (c) The inverse of f is f. Find f ()
17 C3: Functions 7 4 The function f has domain 4 and is defined b f() ( 3) 2. (a) (i) Find the value of f(4) and sketch the graph of f(). (ii) Hence find the range of f. (b) Eplain wh the equation f() has no solution. (c) The inverse function of f is f. Find f (). [A] 5 For each of the following: (i) find the range of the function, (ii) find the inverse function, stating its domain, (iii) state the range of the inverse function. (a) f() (3 ) 3,, (b) g() 2,, (c) h() (2 3) 5, 0, (d) q() 5, 5, 4 (e) r(), For each of the following functions f and g: (i) find the range of the function, (ii) find the inverse function, stating its domain, (iii) state the range of the inverse function. (a) f() 2 5,, 3, 3 (b) g() 5 4,, (a) Sketch the graph of the function f, where f() 2 3,. Eplain wh f does not have an inverse function. (b) Sketch the graph of the function g given b g() 2 3,. Eplain wh g has an inverse and find g (). State the domain and range of g. 8 The function h has domain 0 and is defined b: h() 2 3 (a) Sketch the graph of h() and eplain wh h has an inverse. (b) Find h () and state the domain and range of h. 9 Determine whether an of the functions f, g and h are selfinverse functions. 2 (a) f(), 2, 2, (b) g() 3 4, 2, 2, (c) h() 3 5, 3, 3.
18 8 C3: Functions 0 The function f is defined b f() 2, 2. (a) Sketch the graph of f() and state the range of f. (b) Eplain wh the inverse function f eists and state its domain. Find an epression for f (). [A] The function f has domain 2 and is defined b f() 2 3. (a) Find f(2) and f(00). (b) Determine the range of f. (c) The inverse of f is f. Find f (). 2 The function f with domain 2 is defined b f(). 2 (a) Describe geometricall how the graph of, 0 is transformed into the graph of f(). (b) Sketch the graph of f(). (c) Eplain briefl wh f has an inverse function, state the domain of f, and epress f () in terms of. Worked eamination question The function f has domain 0 2 and is defined b f() 3. (a) Find f(0) and f(2). (b) Sketch the graph of f(). (c) Find the range of f. (d) State, with a reason, whether the inverse function, f, eists. (e) Find ff(), giving our answer in the form 9 a 6 b 3 c. Solution (a) f(0) 0 f(2) 8 9 (b) (2, 9) (0, ) The sketch is not intended to be an accurate plot and so ou should not get too worried about the relative positions of the two endpoints of the graph.
19 C3: Functions 9 (c) The lowest point on the curve is (0, ). The range, therefore, is f() 9. (d) Since the function is oneone, the inverse function does eist. (e) ff() f( 3 ) ( 3 ) Hence, ff() MIXED EXERCISE The function f has domain 2 and is defined b f() 2 5. (a) Find f( ) and f(2). (b) Sketch the graph of f(). (c) Find the range of f. (d) State, with a reason, whether the inverse function, f, eists. (e) Find ff(), giving our answer in the form 4 p 2 q. [A] 2 The functions f and g are defined with their respective domains b 6 f(),, 2 2 g() 2 2,. (a) Find the range of g. (b) The composite function fg is defined for all real values of. Find fg(), giving our answer in the simplest form. (c) The inverse of f is f. Find an epression for f (). (d) The graph of f() and the graph of f () intersect at two points. Find the coordinates of the two points. 3 The functions f and g are defined with their respective domains b 4 f(), 0 3 g() 9 2 2,. (a) Find fg(), giving our answer in its simplest form. (b) (i) Sketch the graph of g(). (ii) Find the range of g. (c) (i) Solve the equation g(). (ii) Eplain wh the function g does not have an inverse. (d) The inverse of f is f. (i) Find f (). (ii) Solve the equation f () f(). [A]
20 20 C3: Functions 4 The function f() with domain { : 0} is defined b 8 f(). 2 (a) Sketch the graph of f and state the range of f. (b) Find f (), where f denotes the inverse of f. (c) Calculate the value of for which f() f (). [A] 5 The functions f and g are defined b f() 3 4, g() 2,, 0. Write down, in a similar form: (a) the composite function fg, (b) the inverse function f. 6 (a) State which of the following graphs, G, G 2 or G 3, does not represent a function. Give a reason for our answer. [A] G G2 G 3 (b) The function f has domain 2 and is defined b f() 5. A sketch of f() is shown opposite. (i) Calculate f(2) and f(0). (ii) Find the range of f. (iii) The inverse of f is f. Find f (). [A] 7 The functions f and g are defined for all real values of b f() 5 3 g() 3 4. (a) Solve the inequalit f(). (b) The composite function fg is defined for all real values of. Find fg(), epressing our answer in the form p q 3, where the values of p and q are to be found. (c) The graph of g() is sketched opposite with equal scales on the  and aes. Cop the graph of g() and, on the same aes, sketch the graph of g (). (d) Find an epression for g (). 2
21 C3: Functions 2 Ke point summar A function is a oneone or a manone mapping. p3 2 The set of numbers for which a function is defined p3 is called the domain. 3 A function f consists of two things: p4 a defining rule such as f() 2 3; its domain. 4 The set of values the function takes for the given p4 domain is called the range. 5 When the domain of f is a continuous interval, the p5 range can be found b considering the graph of f(). The range consists of the possible values that can take. The range of f is written as an inequalit involving f(). 6 The composite function fg means first g then f, since: p0 fg() f[g()]. 7 A function f has an inverse onl when f is oneone. p2 Its graph is obtained b reflecting the graph of f in the line. 8 A reverse flow diagram can be used to find an inverse p3 function when occurs onl once in f(). You consider how f() has been constructed as a sequence of simple operations and set up a flow diagram. Then ou reverse each operation and reverse the direction of the flow to find f (). 9 The inverse of f can be found b the following p4 procedure: Write f(). Rearrange the equation to make the new subject. Interchange and (equivalent to reflecting in ). The new epression for is equal to f ().
22 22 C3: Functions Test ourself The function f is defined for all real values of b Section. f() Find: (a) f(0), (b) f(), (c) f( 2). 2 Find the range of the function f where f is defined b Section.3 f() 3, 3. 3 State the maimum possible domain for the function g where Section.5 g() 4. 4 The functions f and g are defined b f() 2 5 and Section.7 g() 7, and each has domain. Find an epression for gf() in its simplest form. 5 The function f with domain is defined b Section.0 f() (a) Find the range of f. (b) Find the inverse function, f, and state its domain. What to review 6 The function g is defined for all real values of, 4, b Section.0 g() Find the inverse function g and state its domain. Test ourself ANSWERS (a) 2; (b) 0; (c) 0. 2 f() gf() f () domain is all real values of, , (a) f() 9; (b) f () 3 2
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