ACJC 2025 Functions Summary

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Functions / Summary / Page 1 of 1 SUMMARY: Functions Function f Inverse Function f −1 Composite Function gf Existence Each and every element x in the domain is assigned to a unique image  “Vertical line test”, whereby the line xa= , for all f ,aD cuts the graph of f ( )yx= exactly once f must be one-one  “Horizontal line test”: If the horizontal line yk= , for all k , cuts the graph of f ( )yx= at most once , then f is a one -one function. To show one-one: - Draw graph and write “general statement” (above) To show NOT one-one: - Give a counter-example (e.g. from the graph, the line y = 9 cuts the curve f ( )yx= more than once…or f(5) = f(1) = 9…) fgRD Domain Given 1 ffDR− = gf fDD= Rule - Quadratic - Modulus - Rational Functions - Logarithm, Exponential - Trigonometric - Piece-wise 2 for 0 1f( ) 2 for 1 2 xxx xx =− +   - Periodic f ( ) f ( )x k x+= for all real values of x  f is periodic with period k e.g. sin x is periodic with a period of 2π THREE STEPS: 1.Let f ( )yx= 2. 1f ( )xy −= , i.e. Make x the subject - For quadratic, complete the square/use quadratic formula - Choose the correct sign (if needed) based on fD 3.Replace y by x E.g. 2f ( ) 6 14,x x x= − + 3x Let 22f ( ) 6 14 ( 3) 5 y x x x x= = − + = − + 35xy =  − (“  ” important, then we choose based on fD )  since 3x , 35xy= − − ( )1 1 fff ( ) 3 5 , 5 [5, )x x x D R − − = − −  = =   gf ( ) g f ( )xx= (Sub f into g)     2 2 22 2 E.g. f ( ) 5, g( ) sin , gf ( ) g f ( ) g( 5) 5 sin( 5) fg( ) f g( ) f( sin ) ( sin ) 5 x x x x x x x xx x xx xx xx xx = +  = +  = =+ = + + + = =+ = + + Range Graphical Method: 1.Draw f ( )yx= graph and restrict graph to the given domain of f 2.Read off y-values E.g. 2f ( ) 10 ( 2) , 0 5x x x= − −   fR = ( 1,10 Algebraic method: - Discriminant method (Refer to Tut 5, Q7b) 1 ffRD− = To find gfR , Method 1: 1.Sketch the graphs of f and g separately 2.Obtain fR from the graph of f 3.Obtain gfR by reading off range of graph of g using fR as the domain Method 2 (usual method to find range): 1.Sketch the graph of gf 2.Obtain the range of gf from the graph, using gf f()DD= Graphs / Others • Use GC but pay attention to limitations of GC (for example, when sketching ln graphs) • Graphs of f( )yx= and 1f ( )yx −= are reflections of each other about the line yx= . • Point(s) of intersection (if any) of the graphs of f ( )yx= and 1f ( )yx −= can be found by solving: 1f ( ) f ( )xx −= or f ( )xx= or 1f ( )xx− = . • 11ff ( ) f f ( )x x x−− == , however , they are generally different functions as they have different domains: 1 ff f : , x x x D−  while 1 1 fff : , x x x D − −  •   22f ( ) ff ( ) f ( )x x x= • If f f ( ) ,xx= 11f f f( ) f ( )xx−−= 1f( ) f ( )xx −= x y (0,6) (2,10) (5,1)