MATH 149 LABORATORY ASSIGNMENT 5 INVERSE FUNCTIONS Given a function f we wish to define a function g called the inverse of f which reverses the action of f i e whenever f a b then g b a In order for this reversal process to define a function it is necessary that f be one to one for each number b in the range of f there can be only one number a in the domain of f such that f a b If f is one to one no horizontal line can cut the graph of f more than once If g is the inverse of the one to one function f then the graph of g is the set of points f x x x in Dom f In class we showed that Dom g Ran f and Ran g Dom f that g f x x and f g x x that g is one to one with inverse f and that the graph of g is the reflection of the graph of f in the line y x If we differentiate the equation f g x x using the chain rule we obtain the equation f g x g x 1 Solving this equation for g x yields the formula for the derivative of the inverse g x of a function f x g x 1 f g x Now suppose that f a b and therefore that g b a From the last formula it follows that g b 1 f g b 1 f a or g f a 1 f a Since all of the numbers in the domain of g and thus the domain of g are of the form f a for some a in the domain of f it follows that the graph of g is the set of points f a 1 f a a in dom f Maple will use this representation to generate the graph of g Example 1 Somtimes it is easy to find an explicit expression for the inverse g of a given function f by solving the equation f y x for y For example suppose f x 2x 3 3x 7 restart with plots f x 2 x 3 3 x 7 f x 2x 3 3x 7 To find g the inverse of f we interchange x and y and solve the resulting equation for y eq f y x eq 2y 3 x 3y 7 g solve eq y g 3 7x 2 3 x g 3 7x 2 3 x g 3 7 x 2 3 x We will verify the formula for g i e we will show that g x 1 f g x by computing both g x and 1 f g x and showing they are equal Note that we defined g as an expression in the variable x rather than using the Maple variable independent function method this was necessary because we were exchanging the variables x and y To differentiate an expression in the variable x we use the following diff command diff g x 7 3 3 7 x 2 3 x 2 3 x 2 simplify 23 2 3 x 2 Now we compute 1 f g x diff f x x 2 3 2 x 3 3x 7 3 x 7 2 subs x g 2 3 7 x 3 3 2 3 x 2 2 3 3 7 x 3 3 7 x 7 7 2 3 x 2 3 x simplify 2 3 x 2 23 1 23 2 3 x 2 We next plot the graphs of f g and the line y x using the display command found in Maple s plots subpackage with plots A plot f 5 5 5 5 color red discont true B plot g x 5 5 y 5 5 color green discont true j x x j x x C plot j 5 5 color black display A B C Note that the graph of g is the reflection of the graph of f in the line y x Example 2 restart with plots x f x x 2 sin 4 The inverse of f will again be denoted by g Consider the function 0 x 8 In this case as we will see below we are not able to explicitly solve the equation f y x for y g x and thus generate a formula for g x However we can still plot the graph of the inverse g and its derivative g and compute their values for numbers in their domains We first try to find a formula for g by solving f y x as in example 1 f x x 2 sin Pi x 4 1 f x x 2 sin x 4 eq f y x y x eq y 2 sin 4 solve eq y 4 RootOf 4 Z sin Z 2 x Apparently Maple cannot solve this equation for y g x in terms of x However even though we don t have a formula for g we can still easily generate the graph of g since we know it consists of the set of points of the form f x x for x in dom f Note the use of the textplot command from the plots package which allows us to label the graphs The first two entries after the in the command are the X and Y coordinates of the point in the graph where the label begins For more information see plots textplot in Help fGraph plot x f x x 0 8 color red gGraph plot f x x x 0 8 color green identity plot x x x 1 5 7 color black T1 textplot 2 8 6 6 y g x T2 textplot 6 6 2 4 y f x display fGraph gGraph identity T1 T2 scaling constrained Next we plot g and g on the same set of axes Recall that the graph of g is the set of points f x 1 f x x in dom f fprime x D f x fprime x D f x DgGraph plot f x 1 fprime x x 0 8 color red T3 textplot 4 2 95 y g x display gGraph DgGraph T1 T3 scaling constrained Recall that g x 1 f g x 2 x 6 gprime 1 fprime g x gprime 1 1 1 1 cos g x 4 4 For example we may find g 2 by utilizing the fact that f 4 2 implies g 2 4 subs g x 4 x 2 gprime 1 1 1 cos 4 value 1 1 4 evalf 4 659792369
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