RI Functions Solns

RAFFLES INSTITUTION H2 Mathematics 9758 2023 Year 6 Term 3 Revision 10 (Summary and Tutorial) Topic: Functions Summary for Functions Relations and Functions  The set of inputs of a function is known as the domain while the corresponding set of outputs is known as the range.  To define a function completely, both the rule and domain must be specified.  The range of a function can be obtaining by looking at its graph (y-values).  If a function is such that no two elements in the given domain have the same output, it is a one- one function.  To check if a function is one-one, we use horizontal lines to check. If function is one-one, such as 2f: , 2xx x  , we write since every horizontal line y = k,  4,k , cuts the graph of f at one and only one point, f is a one-one function. Note that  4, is the range of f. If function is NOT one-one, such as 2f: ,xx x  , we choose a specific line y = k where k is a value in range of f. In this case, we can write since the horizontal line y = 1 cuts the graph of f at more than one point, f is not a one-one function Inverse Functions  Given a function f, the inverse function 1f exists if and only if f is one-one.  To define 1f , you must find the rule and domain.  To find the rule of 1f , let f( )yx and make x the subject, i.e -1f()xy . Example to find the rule of inverse of g: 1 e , xxx  Let 1e xy  and make x the subject to get ln( 1)xy  .  To find domain of 1f , use the result: 1fD   fR Note also that 1 ffRD  .  The graphs of f( )yx and 1f( )yx  are reflections of each other in the line yx .

Composite Functions  Given 2 functions f and g, the composite function gf exists if and only if fRD g .  To define gf, you must find the rule and domain.  To find the rule of gf, find f( )x first, then find gf () x .  To find domain of gf, use the result: gfD  fD  In general, fg gf . i.e. composition of functions is not commutative.  The composite function ff, if it exists, is written as 2f , i.e.  2f ( )f f ( )f f ( )xx x . o Useful results: (i) 1ff ( )xx  , where 1fDx  . (ii) 1ff ( ) xx  , where fDx .  To find the range of gf, use either ( 1 ) gfR  gR restricted to fR or (2) sketch the graph of gf ( )yx over gfD and find gfR . Example Find the range of the composite gf given that the functions f and g are defined as follows: f : 2, 3xx x  , g : , 0xxx  . Method 1 Sketch the graphs of the functions f and g on separate diagrams. For the composite function gf, f is first applied to each fDx to find the image f( )x , and then the rule for g is applied to f( )x to obtain the image gf ()x . i.e. g f () gf ()xx . Consequently, we can obtain the range of gf as the range of g whose domain is restricted range of f, i.e gf[ 3 ,) [ 1 ,) [ 1 ,)    . Thus  gfR1 , . x  y 4 2 O 3 1 2 1 Rf x  y 2 O 1 1 Rgf

Method 2 Find the function gf followed by sketching its graph to find the range.   gf fgf ( ) g f ( ) 2, D D 3,xx x     From sketch,  gfR1 , .   Rgf

Revision Tutorial Questions Source of Question: EJC/Y5 Promo/2018/Q3 1 The function f is defined by f: l n1 2 , , 1 2xx x x   . (i) Explain why f does not have an inverse. [1] The function g is now defined by g: l n1 2 , xx x a  where a is a real constant. (ii) State the minimum value of a such that 1g exists. [1] (iii) Using the value of a stated in (ii), define 1g in a similar form and state the range of 1g . [4] [ (ii) 1 2a (iii)  1 1g: e 1 , 2 xxx  ; 1g 1R, 2   ] Solution: 1(i) [1] From sketch, the line 1y intersects the graph of f( )yx twice. Hence, f is not one-one and does not have an inverse. Alternatively, since f( 0 ) 0 f( 1 ) , f is not one-one and does not have an inverse. 1(ii) [1] Minimum value of 1 2a 1(iii) [4] Let ln 1 2yx For 1 2x , ln(2 1)yx 21 e yx 1 e12 yx  1 1g( ) e 1 2 yy  , 1 ggDR    1 1g: e 1 , 2 xxx   and 1 gg 1RD , 2    x  y O 1 1y    f( )yx

Source of Question: NJC/Y5 Promo/2018/Q10 2 The functions g and h are defined by 2 8g: 2 xx x  , xk ,  21h: 2 4xx  , 04 x . (i) Find the smallest exact value of k such that 1g exists. [2] Use 2k  for the remainder of this question. (ii) Find the exact range of g. [1] (iii) Sketch the graphs of gyx , 1g( )yx  and 1gg ( )yx  on the same diagram. Hence find the exact solution of  1g( ) g .xx  [6] (iv) Show that hg exists and determine the range of hg. [3] [ (i) 2k  (ii) 80, 3    (iii) 6x (iv)  0, 1 ] Solution: 2(i) [2]  2 8g 2 xx x       2 2 2222 82 8 2 16 8g 22 xx x xx xx     For maximum point of gyx , g0 x  . 216 8 0 2 x x   From the graph, it is a one-one function if 2x , so the smallest exact value of k is 2 . 2(ii) [1]  2 16 8g2 22 3  , so the range of g is 80, 3    as observed from the graph in (i).

2(iii) [6] From the diagram, we may solve g xx instead of  1g( ) g .xx     2 3 3 2 8 2 82 60 60 66 0 x xx xx x xx xx xx x       0x or 6x or 6x From the diagram in (ii), 6x . 2(iv) [3] Since gh 8R0 , 0 , 4 D3    , hg exists.   h80, 0, 13                  x y     y x O Graph of h 1gg (y x

Source of Question: MI/Promo/PU2/2017/Q6 3 A function f is defined by    2 f4 3x x for .x  (i) Show that 1f does not exist. [1] (ii) If the domain of f is restricted to x k , state the largest exact value of k for which the function 1f exists. [1] (iii) Using the domain defined in part (ii), find 1f and its range. [3] (iv) Solve the equation     1ff x x exactly. [3] [(ii) 3k  (iii)     1f3 4 ,, 4xx x      ; 1fR, 3   (iv) 55 2x  ] Solution: 3(i) [1] Since the horizontal line 1y intersects the graph of  fyx more than once, f is not one-one. Hence, f −1 does not exist.    Alternatively, since f( 1 ) 0 f( 4 ) , f is not one-one and does not have an inverse. 3(ii) [1] 3k  3(iii) [3] Let  2 43y x 34 x y   34x y  Since 3x , 34x y  Hence,     1f3 4 ,, 4 .xx x      So 1 ffRD , 3    3(iv) [3] We may solve  f x x instead of     1ff x x .   2 2 43 55 0 x x x x       52 5 4 ( 5 ) 55 or (rejected 3)5 2 5 2 2xx        1y

Source of Question: NYJC/Prelim/2017/01/Q7 4 The functions f and g are defined by 2 f: e , , 0 xxx x    , 1g: , , 3 3xx x x    . (i) Show that 1g exists, and define 1g in a similar form. [2] (ii) State the solution set for  1gg xx  . [1] (iii) Explain why 1fg does not exist. [1] Let the function h be defined by h: g , , xx x x k   , where k is a real constant. (iv) Given that 1fh  exists, state the maximum value of k . [1] (v) For the value of k found in (iv), (a) find the exact range of 1fh  , [2] (b) solve   1hhxx  . [2] [(i) 1 1g: 3 , , 0xx x x     (ii)  |0xx (iv) 3k  (v)(a) 1 9 fhR0 , e  (v)(b) 3.30 x  ] Solution: 4(i) [2] (i) Every horizontal line yk cuts the graph at most once. This implies g is one-one. Therefore 1g exists

1 1g: 3 , , 0xx x x     4(ii) [1]  |0xx 4(iii) [1]  1 ggRD \ 3   , fD 