00010001010008main.ACT000102000flogarithmic.EAC010000009937 y=x^(2)+2``vrALCuxyH ! x^2+2^[About the shiftings of[logarithmic functions.[\AuthorProfessor Dr. Wei-Chi Yang Department of Math/Stats Radford University, Radford, VA 24142 USA e-mail: wyang@radford.edu URL:www.radford.edu/ ~wyang[ Objective:[(1) Traditionally, students[do not have so many [problems to find the [inverses of functions [such as f(x)=ln(x+a)+b; [but we will investigate[ further. [(2) We shall investigate [the effects of the [coefficients a and b on [the graphs of [f(x)=ln(x+a)+b[and its respective [ inverses.[(3) We extend the [observations above to an[arbitruary function and[its inverse (if the [inverse exists).[[Example.[Investigate how the [coefficient 'a' affect[the graphs of y=ln(x+a) [and the its inverse. [Step 1.[First, we find the inverse[ of y=ln(x+a).R clear_a_zRdoneRsolve(y=ln(x+a),x)Rx=-a+yR[Thus, the inverse of[f(x)=ln(x+a) is [g(x)=x-a.[-> We check if [f(g(x))=g(f(x))=x below.[(Be sure the set the[format to be 'Decimal[ Calculations'[under settings.)Rdefine f(x)=ln(x+a)RdoneR define g(x)=x-aRdoneRf(g(x))RlnxRg(f(x))Rx[Step 2.[We will investigate how[y=ln(x+a) is related to[y=ln(x)[\->Open the graph LGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10Hy2HHy3dH              ( 4  @ !L "X #d $p %| & ' ( ) * + , - . 0 1 2 3 4$ 50 E< FH HT I` Jl Kx L M N O P Q R S T ] ^ _ ` a b    , 8 system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3x_(x)  3x_(x+5)  3x_(x-5)`P`P`p`p `(10qy`#YuY ` R[Remark: We see that [y=ln(x+5) and y=ln(x-5) [are horizontal shiftings[of y=ln(x) to the[left 5 units and right 5[units respectively.[[Step3.[We will investigate how[y=x-a is related to [y=x.[\->Open the graph PGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10Hy2LHy3hH              , 8  D !P "\ #h $t % & ' ( ) * + , - . 0 1 2 3 4( 54 E@ FL HX Id Jp K| L M N O P Q R S T ] ^ _ ` a b   $ 0 < system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $   3xexp(x)  3xexp(x)+3  3xexp(x)-3`P`P`p`p `(10qy`#YuY ` R[Remark: [(1) We see that[y=x +3 and y=x-3 are['vertical' shiftings of [y=x up 3 units and [down 3 units respectively.[(2) Consequently, if[y=ln(x) moves to the [ left 5 units [(so the equation becomes [y=ln(x+5)), the [correponding inverse[(y=x-5) will[moves down 5 units.[(3) Similarly, if[y=ln(x) moves to the [right 5 units [(so the equation becomes [y=ln(x-5)), the [correponding inverse[(y= x+5) will[moves up 5 units.[[Now we shall use the[animation to see the[graphs of y=ln(x+a) and[y=x-a.\ Exploration 1 HGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10Hy2LHy3hH|              $ 0  < !H "T #` $l %x & ' ( ) * + , - . 0 1 2 3 4 5, E8 FD HP I\ Jh Kt L M N O P Q R S T ] ^ _ ` a b    ( 4 system]listsystem^]=system_^system`_systema`systemb~ a^x-axb system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $   3x_(x+a)  3x^x-a 3xx`P`P`` x)) `xx(10qy`x#YuYx `` ``cxsh R[->Open the the graph[editor above and tap on[ 'graph' icon.[->Ignore the 'error' [message.[->Select the diamond [ shape icon.[->Select 'Modify' and[->Tap on the graph we[observe the followings:[[Remark:[This confirms our previous[observations that if a[function f is being shifted[to the left 'a' unit(s), [then the corresponding[inverse will be shifted[down 'a' units and so on.[Exploration 2.[There is another [interesting way of [finding the inverse[of a function. First open[the graph editor below.\Type a function  Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1Hy28HT ` l x               ! ", #8 $D %P &\ 'h (t ) * + , - . 0 1 2 3 4 5 E F H( I4 J@ KL LX Md Np O| P Q R S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $   3x_(x-2)  3xexp(x)+2`P`P`p`p `(10qy`#YuY 0% )`` Q[->When a function is[ selected.[->Plot the graph and[go to 'Analysis' icon[->Sketch->Inverse. We [get the inverse of the [ function.[We can guess (by now)[what the inverse should[be.[ Exercise 1.[(1) Investigage how the [variable 'a' will affect[the graph of y=x3+a[and its inverse.[(2) Investigage how the [variable 'a' will affect[the graph of y=(x+a)3[and its inverse.[ Solution:[(1) We first define the[function and find its [inverse below.R clear_a_zRdoneR define f(x)=x3+aRdone[**Basic Format should be[set 'Complex' to get [a symbolic expression.Rsolve(y=x3+a,x)Rx=y-a13,x=-y-a132-3y-a132,x=-y-a132+3y-a132Rdefine g(x)=(x-a)13RdoneRf(g(x))Rx[**Switch the 'Basic[Format' back to 'Decemal'.[Rg(f(x))Rx[Therefore, f and g are[ inverses.\The variable a  Graph2DT Graph3Dh LISTSYSt4NModify $ STATCALC NSTATSYS \NSequence0, Sheet\| Sheet3D| SolveEqTSolveLwrX SolveUprd StupFLG1p(StupListD StupPict$ ViewWind y1Hy24 HT ` l x               ! ", #8 $D %P &\ 'h (t ) * + , - . 0 1 2 3 4 5 E F H( I4 J@ KL LX Md Np O| P Q R S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ ax-a)^(1/b  system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3xx^3+a 3x(x-a)^(1/3)`P`P`p`p `(10qy`#YuY `` Q[Remark:[(1) We see that the [graph of y=x^3+a is [a vertical shifting of [y=x^3.[(2) Specifically, when [y=x^3 moves up 5 units[say, y=x^3+5, the[inverse becomes [y=(x-5)13 , which is a [horizontal shifting of['right 5' from y=x13.[(2) The solution to this[one should be obvious.[The graph of y=(x+a)3 is[the horizontal shiftings[of y=x3 (If a<0, then[it is shifted to the right['a' unit(s), if a>0, then[it is shifted to the left[ 'a' units.[Consequently, the inverse[ of y=(x+a)3 should be[shifted vertically up 'a'[units(if a<0) and[down 'a' units (if a>0).[[ Exercise 2.[(1) Investigage how the [variable 'a' will affect[the graph of y=x2+a, [x>0 and its inverse.[(2) Investigage how the [variable 'a' will affect[the graph of y=(x+a)2,[where x>-a, and its inverse.[[Hints:[(1) We first sketch the[ graph of y=x2+a with [the following geometry[link; replace 'a' with [some positive and negative[numbers and see how 'a'[affects the graph.]\A vertical shifting[(2) Now we will see[if y=x2+a has an inverse[by using the following[ graph editor:\Inverse of y=x^2DGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10Hy2HHy3`Hx               ,  8 !D "P #\ $h %t & ' ( ) * + , - . 0 1 2 3 4 5( E4 F@ HL IX Jd Kp L| M N O P Q R S T ] ^ _ ` a b    $ 0 system]listsystem^]=system_^system`_systema`systemb~tttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt aseq_histbNewFolde system]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemtsystem]listsystem^]=systemt  0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D 0@PSheet1ju^Sheet2ju^Sheet3ju^Sheet4ju^Sheet5ju^SheetSheet3D i ^ $  3xx^2 3xx^2+1 3xx^2-1`P`P`p`p `(10qy`#YuY `a` R[-After graphing y=x^2,[we select Analysis->[Sketch->Inverse. We get[a horizontal parobola,[x=y^2; which is not a [ function.[-Therefore, if we [restrict the domain for[f(x)=x^2 to be x>0, [then the corresponding[inverse will be[ f-1(x)=x.[We leave readers to [verify if f and  f-1 are[inverses graphically and[algebraically.[[(3) Find the inverses for[f(x)=x^2+a with various[ number a.eActa `b fH xx^(3)+agH$x(x-a)^(((1)/(3)))