INVERSE FUNCTIONS

created by :  Eric Hofelich

E-Mail address  ->  ejhofelich@kc.il.us

Mathematics Instructor @ Kaskaskia College

Centralia, Illinois

*This module has been created for students enrolled in College Algebra that are seeking further insight of Inverse Functions.

To open one of the sections below, simply  double-click on the harpoon shaped arrow to the far right of each section. 
	To close a section, double-click on the bracket to the right that "groups" the cells in that section.

initialization

Please execute the following cell before continuing!!!!

<< "Graphics`Colors`"

Introduction

First lets look at the plot of the graph containing the following ordered pairs:  (-2,1), (0,2), (2,5), (4,9)

\(f = {{\(-2\), 1}, {0, 2}, \ {2, 5}, \ {4, 9}}; 

    fplot = ListPlot[f, PlotStyle -> {PointSize[.04], Red}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        PlotRange -> {{\(-10\), 10}, {\(-10\), 10}}]; \)

By interchanging the first and second coordinates of each ordered pair, you can form the inverse function of f.  This function is denoted by (f^(-1)) .
f     = (-2,1), (0,2), (2,5), (4,9)
(f^(-1)) = (1,-2), (2,0), (5,2), (9,4)
                                                            Lets look at the plot of (f^(-1))

\(\(	inv = {{1, \(-2\)}, {2, 0}, {5, 2}, {9, 4}}; 

    newinv = ListPlot[inv, PlotStyle -> {PointSize[.04], Red}, 
        AspectRatio -> Automatic, 
        PlotRange -> {{\(-10\), 10}, {\(-10\), 10}}]; \)\)

Now let's look at the graphs of f and (f^(-1))on the same coordinate axes

\(yline = 
      Plot[x, {x, \(-5\), 10}, PlotStyle -> RGBColor[0, 0, 1], 
        AspectRatio -> Automatic, DisplayFunction -> Identity]; 

    flabel = Graphics[Text["f", {5, 10}]]; 

    invlabel = Graphics[Text[\*""\!\(f\^\(-1\)\)"", {10, 3}]]; 

    symlabel = Graphics[Text["y=x", {10, 9}]]; 

    Show[{fplot, newinv, yline, flabel, invlabel, symlabel}, 
      DisplayFunction -> $DisplayFunction]; \)

Notice that f and its inverse function (f^(-1))are symmetrical about the line y=x , that is ,f and (f^(-1))are mirror images of each other with respect to the line y=x.

The domain of f is equal to the range of (f^(-1)), and vice versa

Definition of Inverse Function:
 A function (f^(-1))is the inverse of  the function f if f((f^(-1))(x)) = x for each x in the domain of (f^(-1))and (f^(-1))(f(x)) = x for each x in the domain of f.

Problem 1

Given:  f(x) = 5x + 1 and  (f^(-1))(x) = ((x  -  1)/5)

a) What is the domain and range of f(x)?

b) Find f(-2), f(-1), f(0), and f(1)

c)  Using the points found in part b), plot the graphs of f (x) and (f^(-1))(x) on the same coordinate axes.  Also, make sure to include the graph of y=x to show the symmetry.

d)  What is the domain and range of (f^(-1))(x)?

e)  Verify that f and (f^(-1))are inverses of each other by showing that f((f^(-1))(x)) = x and (f^(-1))(f(x)) = x

Graphs of the Inverse of a Function

Looking at the graph of an inverse function is an excellent way of getting deeper insight into the concepts of symmetry, domain, and range of functions. It also provides a better grasp of the concept of an inverse function.  Remember that domain represents your x values that can be used in a function and that range represents your y values of a function (ie.  the values of f(x)).

Consider the graph below!!!

\(firstplot = 
      Plot[x\^2, {x, 0, 3}, 
        PlotStyle -> {Thickness[ .02], RGBColor[1, 0, 0]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    secplot = 
      Plot[\@\(x\ \), {x, 0, 8}, 
        PlotStyle -> {Thickness[ .02], RGBColor[0, 0, 1]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    thirdplot = 
      Plot[x, {x, 0, 8}, 
        PlotStyle -> {Thickness[ .001], RGBColor[0, 1, 0]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    labelf = Graphics[Text[\*""f(x)=\!\(x\^2\)"", {4, 8}]]; 

    labelinv = 
      Graphics[Text[\*""\!\(f\^\(-1\)\)(x)=\!\(\@x\)"", {8, 2}]]; 

    labelaxes = Graphics[Text["y=x", {8, 7}]]; 

    Show[{firstplot, secplot, thirdplot, labelf, labelinv, labelaxes}, 
      DisplayFunction -> $DisplayFunction]; \)

The domain of f(x) is [0,∞) while the range of f(x) is [0,∞).  This is also true for it's inverse, (f^(-1))(x).
Why is the domain for (f^(-1))(x) from [0,∞)?  Why is the domain for f(x) from only [0,∞)?

The answers to these questions are very important to the existence of an inverse function.  The domain of (f^(-1))(x) will always be from the range of f(x) and visa versa .  Also, we know that we could not take the square root of negative numbers.  The domain of f(x) is restricted to only the non-negative reals because f(x) must be one to one for the inverse to exist.

Let's look at this in better detail!!            We know the graph of f(x) = (x^2)is a parabola.  But, a parabola is not one to one and therefore would not have an inverse.  It would, however, have an inverse if we restricted our domain as above. Let's look at the algebra involved in finding the inverse of f(x).

					f(x) = (x^2)           domain -> [0,∞)
					  y   = (x^2)          replace f(x) with y
					  x   = (y^2)          interchange x and y
					  y   =  ± (Sqrt[x])    square root of both sides
				       (f^(-1))(x)  =   ± (Sqrt[x])   replace y with inverse notation
				       (f^(-1))(x)  =  +(Sqrt[x])     since domain of f(x) [0,∞)

If we restricted our domain of f(x) to the negative reals including zero (-∞,0] our inverse function would be (f^(-1))(x) = -(Sqrt[x]).
Thus our graph would change as seen below;

\(firstplot = 
      Plot[x\^2, {x, \(-3\), 0}, 
        PlotStyle -> {Thickness[ .02], RGBColor[1, 0, 0]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    secplot = 
      Plot[\(-\@\(x\ \)\), {x, 0, 8}, 
        PlotStyle -> {Thickness[ .02], RGBColor[0, 0, 1]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    thirdplot = 
      Plot[x, {x, \(-4\), 8}, 
        PlotStyle -> {Thickness[ .001], RGBColor[0, 1, 0]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    labelf = Graphics[Text[\*""f(x)=\!\(x\^2\)"", {\(-1.5\), 8}]]; 

    labelinv = 
      Graphics[Text[
          \*""\!\(f\^\(-1\)\)(x)=-\!\(\@x\)"", {7, \(-1.8\)}]]; 

    labelaxes = Graphics[Text["y=x", {8, 7}]]; 

    Show[{firstplot, secplot, thirdplot, labelf, labelinv, labelaxes}, 
      DisplayFunction -> $DisplayFunction]; \)

f(x)						(f^(-1))(x)
                   
Domain                    (-∞,0]                                                         [0,∞)
Range                       [0,∞)                                                          (-∞,0]

Notice that the domain of f(x) is the range of (f^(-1))(x) and the range of f(x) is the domain of (f^(-1))(x).

Problem 2

Consider the graph below to answer the following questions.

\(ggplot = 
      Plot[\@\(2  x - 3\), {x, 1.5, 5}, 
        PlotStyle -> {Thickness[ .02], RGBColor[1, 0, 0]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    ginvplot = 
      Plot[\(x\^2 + 3\)\/2, {x, 0, 3}, 
        PlotStyle -> {Thickness[ .02], RGBColor[0, 0, 1]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    symplot = 
      Plot[x, {x, \(- .5\), 5}, 
        PlotStyle -> {Thickness[ .001], RGBColor[0, 1, 0]}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; 

    gglab = Graphics[Text[\*""f(x)=\!\(\@\(2  x - 3\)\)"", {3, 6.2}]]; 
    
ginvlab = 
      Graphics[Text[
          \*""\!\(f\^\(-1\)\)(x)=\!\(\(x\^2 + 3\)\/2\)"", {4.5, 1.5}]]; 
    
symlab = Graphics[Text["y=x", {5, 4}]]; 

    Show[{ggplot, ginvplot, symplot, gglab, ginvlab, symlab}, 
      DisplayFunction -> $DisplayFunction]; \)

a).  The graph above is of f(x) = (Sqrt[2  x - 3]) and its inverse (f^(-1))(x) = (((x^2)  +  3)/2).  Find the domain and range of each function.
b).  Beginning with f(x), show the algebraic steps needed to find the equation of the inverse of f(x).
c).  Verify that the two functions are inverses of each other by showing that f((f^(-1))(x)) = (f^(-1))(f(x)) = x

Problem 3

i.  Find (f^(-1))(x)  if      a).  f(x) = (Sqrt[(x^2)  - 1]) for x ≤ -1
                              b).  f(x) = (x^3) + 1
ii.  Graph both f and (f^(-1)) on the same coordinate plane for each part above.

The Inverse of an Exponential Function

Consider the graph of the exponential function f(x) = (2^X)

Plot[(2^x), {x, -5, 5}, AspectRatio -> 1, 
      AxesLabel -> {"x", "y"}, 
      PlotStyle -> {Thickness[.02], Red}];

This function is one to one since it would pass the horizontal line test (ie. for each y value there is only one x value).  Therefore it must have an inverse. Let's exam the graph a little more closely.

f(x) = (2^(X  ))

f(-2) = (2^(-2))= 0.25             f(2) = (2^2) = 4
f(-1) = (2^(-1))= 0.5               f(3) = (2^3) = 8
f(0)  = (2^0) = 1
f(1)  = (2^1) = 2

expgraph = 
      ListPlot[{{-2, .25}, {-1, .5}, {0, 1}, {1, 2}, {2, 4}, {3, 8}}, 
        PlotStyle -> {PointSize[.03], Blue}];

To find the inverse of f(x) let us first interchange the x and y coordinates and plot these points.

\(switchpts
       = {{.25, \(-2\)}, {.5, \(-1\)}, {1, 0}, {2, 1}, {4, 2}, {8, 3}}; 

    plotinv = ListPlot[switchpts, PlotStyle -> {PointSize[.03], Red}]; 
    \)

Now, let's combine the two graphs to see the beautiful symmetry about the line y=x.

symline = 
      Plot[x, {x, -2, 8}, PlotStyle -> Green, 
        DisplayFunction -> Identity]Show[{expgraph, plotinv, symline}, 
      AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
      DisplayFunction -> $DisplayFunction];

Finally, let's look at the graphs with the points connected.

\(bluplot = 
      Plot[2\^x, {x, \(-3\), 3}, PlotStyle -> {Thickness[0.03], Blue}, 
        DisplayFunction -> Identity]; 

    redplot = 
      Plot[Log[2, x], {x, .09, 8}, PlotStyle -> {Thickness[0.03], Red}, 
        DisplayFunction -> Identity]; 

    blulabel = Graphics[Text[\*""f(x) = \!\(2\^X\)"", {5, 8}]]; 

    grelabel = Graphics[Text["y=x", {8, 7}]]; 

    redlabel = 
      Graphics[Text[\*""\!\(f\^\(-1\)\)(x)=log[2,x]"", {6, 1}]]; 

    Show[{bluplot, redplot, blulabel, grelabel, redlabel, symline}, 
      AspectRatio -> Automatic, AxesLabel -> {"x", "y"}, 
      DisplayFunction -> $DisplayFunction]; \)

The inverse function (f^(-1))(x) found is called a logarithm, in this case of base 2.

Problem 4

a) Plot the graph of f(x) = (3^X) by plotting points using the following values for x; 
                                                                                                 x = -2,-2,0,1,2 
b) Now, plot the graph of (f^(-1))(x) on the same coordinate system as in a)
c) Write the equation of (f^(-1))(x).
d) What is the domain and range of each function?