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IIT MATH 149 - MATH 149 LABORATORY ASSIGNMENT 5

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MATH 149LABORATORY ASSIGNMENT 5INVERSE 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 ) = − 2 x 3 + 3 x 7 .> restart;> with(plots):Warning, the name changecoords has been redefined > f:=x->(2*x-3)/(3*x+7); := f → x − 2 x 3 + 3 x 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 = − 2 y 3 + 3 y 7x> g:=solve(eq,y); := g− + 37x− + 23x> g := -(3+7*x)/(-2+3*x); := g− + 37x− + 23x 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 weuse the following "diff" command:> diff(g,x);− + 7− + 23x3( ) + 37x()− + 23x2> simplify(%);23()− + 23x2Now we compute 1/ f '(g (x) ):> diff(f(x),x); − 2 + 3x73( ) − 2x3() + 3 x 72> subs(x=g,%); − 2− + 3( ) + 37x− + 23x73⎛⎝⎜⎜⎞⎠⎟⎟− − 2( ) + 37x− + 23x3⎛⎝⎜⎜⎞⎠⎟⎟− + 3( ) + 37x− + 23x72> simplify(%);()− + 23x223> 1/(%);23()− + 23x2 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 → xx> 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):Warning, the name changecoords has been redefined Consider the function f ( x ) = − + x 2⎛⎝⎜⎜⎞⎠⎟⎟sinπ x4 ; 0 < x < 8 . The inverse of f will again be denoted by g. 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); := f → x − + x 2⎛⎝⎜⎜⎞⎠⎟⎟sin14πx > eq:=f(y)=x; := eq = − + y 2⎛⎝⎜⎜⎞⎠⎟⎟sinπ y4x> solve(eq,y);4( )RootOf + − − 4 _Z π ()sin _Z 2 ππxπ 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});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 fx> 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});Recall that g '(x) = 1/f '(g(x)) , -2 < x < 6 . > gprime:=1/fprime(g(x)); := gprime1 + 114⎛⎝⎜⎜⎞⎠⎟⎟cos14π ()g x π 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 + 114()cos ππ> value(%);1 − 1π4> evalf(%);4.659792369> > restart;_______________________________________________________________________Name: <your name goes here>Lab 5Section: <your section goe here> Exercises For the functions specified in (i), and (ii) below, let g denote f inverse. Then: (a) Plot a graph of f, g, and the identity function j ( x ) = x in the same coordinate plane. Include in your plot a label of the two curves: y = f ( x ) and y = g ( x ) .> > (b) Plot a graph of g and g' in the same coordinate plane .


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IIT MATH 149 - MATH 149 LABORATORY ASSIGNMENT 5

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