2'8 Combining Functions

1
2.8 Combining Functions

• Assume we have two functions, f and g
• or f(x) and g(x)
• assume the domain for f(x) is A and the domain
  for g(x) is B
• we can combine these functions in several ways
• f g f(x) g(x) also written as (f g)(x)
• f g f(x) g(x) also written as (f g)(x)
• f g f(x) g(x) also written as (f g)(x)
• f / g f(x) / g(x) also written as (f / g)(x)
• the domain for the above functions is A
  intersected with B
• except for f / g whose domain is A intersected
  with B such that g(x) lt gt 0

2
Examples

• Let f(x) x3
• And g(x) x1/2
• f g x3 x1/2
• f g x3 x1/2
• f g x3 x1/2 x3/2
• f / g x3 / x1/2 x6
• the domain for each of the above is x gt 0 except
  for f / g whose domain is x gt 0
• (f g)(4) 43 41/2 66
• (f g)(4) 62
• (f g)(4) 128
• (f / g)(4) 46 2048

• If functions have differing domains, the
  combination of functions will have a domain that
  is the intersection of domains
• but the domain must also exclude g(x) 0 if g(x)
  is in the denominator for f / g
• Let f(x) x1/2
• And g(x) (4 x2)1/2
• f(x) has a domain of x gt 0
• g(x) has a domain of -2, 2
• f g, f g and f g all have domains of 0, 2
• f / g has a domain of 0, 2)

3
Example Addition Via Graphs

• When two functions are added, the graph of the
  added functions is basically the sum of the two
  y-coordinates of each graph
• On the right, we have the graphs
• f(x) sqrt(x)
• g(x) x 2
• fg sqrt(x) x 2
• At x 0, we have 0 0 2 2
• At x 1, we have 1 1 2 4
• At x 2, we have 2 sqrt(2) 2 which is about
  5.41
• See figure 1 on page 220 for another example

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Compositions of Functions

• A composition of functions is placing one
  function inside another by using one function for
  x
• for instance, if we have f(x) and g(x), we can
  create a composition of f(g(x))
• example f(x) square root(x) and g(x) x2 1
• f(g(x)) square root(x2 1)
• g(f(x)) x(square root x)2 1 x 1
• We often denote a composition of functions using
  o as in f o g(x) f(g(x))
• the domain of a composition f(g(x)) is the domain
  of g(x) that can be in the domain of f(x)
• if some value q is in the domain of g(x) and it
  can be in the domain for f(x), then q is in the
  domain of f(g(x))
• if some value r is in the domain of g(x) but is
  not in the domain of f(x), then r is not in the
  domain of f(g(x))
• if some value s is not in the domain of g(x) then
  s is not in the domain of f(g(x))

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Examples

• Let f(x) x2 and g(x) x 3
• Find f o g and g o f and their domains
• f(g(x)) (x 3)2 x2 6x 9 and the domain
  is all real numbers
• g(f(x) x2 3 and the domain is all real
  numbers
• Find (f o g)(5) and (g o f)(7)
• f(g(5)) f(5 3) f(2) 22 4
• g(f(7)) g(49) 46
• Let f(x) (x)1/2 and g(x) (2 x)1/2
• Find f o g, g o f, f o f, g o g
• f(g(x)) ((2 x)1/2)1/2 (2 x)1/4
• g(f(x)) (2 (x)1/2)1/2)
• f(f(x)) ((x)1/2)1/2) x1/4
• g(g(x)) (2 - (2 x)1/2)1/2
• Let f(x) x / (x 1), g(x) x10, and h(x) x
  3
• Find f o g o h
• f(g(h(x))) f(g(x 3)) f((x3)10) (x 3)10
  / ((x 3)10 1)

6
Recognizing A Composition

• Recall from 2.5, we saw how to graph a function
  that was the transformation of another function
• such as f(x) (x 7)2 is a transformation of
  g(x) x2
• We can similarly detect composition of functions
• Let F(x) (x 9)1/4
• find f(x) and g(x) such that F(x) f(g(x))
• f(x) x1/4
• g(x) x 9
• f(g(x)) (x 9)1/4 F(x)
• Example a ship is traveling 20 miles per hour
  parallel to a straight shoreline and is 5 miles
  form shore, and passes a lighthouse at noon
• express the distance s between the lighthouse and
  the ship as a function of d
• s f(d) (25 d2)1/2
• express d as a function of t
• d g(t) 20t
• find f o g
• f(g(t)) ((25 (20t)2)1/2

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2.9 One-to-One Functions and Inverses

• A function is one-to-one if for every element in
  the functions domain x1 and x2, f(x1) does not
  equal f(x2)
• that is, the function will not repeat any values
• recall that a function can not generate 2 values
  for a given x, now it cannot produce a repeat
  value for different xs
• the result will be that a graph of any one-to-one
  function will pass a horizontal line test any
  horizontal line drawn through the graph will
  cross the graph at most in one place (this is
  similar to the vertical line test)
• The importance of a one-to-one function is that
  it can have an inverse function
• if we have a function f(x) which produces a value
  y then the inverse function f(y) x

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Examples

• Is f(x) x3 a one-to-one function?
• Unlike squares, no two numbers will have the same
  cube, so f(x1) does not equal f(x2) for any x1
  not equal to x2
• So, f(x) is a one-to-one function
• We can also determine this by graphing the
  function (see to the right) and see that it
  passes the horizontal line test
• Is g(x) x2 a one-to-one function?
• Since g(-1) 1 g(1), no it is not
• We can also see that it does not pass the
  horizontal line test

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Showing that a Function is One-to-One

• Show that f(x) 3x 4 is one-to-one
• Suppose we have two numbers, x1 and x2
• f(x1) 3x1 4
• f(x2) 3x2 4
• for f(x1) to equal f(x2) then
• 3x1 4 3x2 4 or 3x1 3x2 or x1 x2
• So, for f(x) to equal f(y), x must equal y
• Therefore, f is one-to-one

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Inverse of a Function

• We denote the inverse of a function f(x) as
  f-1(x)
• A function and its inverse has the following
  properties
• f-1(f(x)) x for every x in fs domain
• f(f-1(x)) x for every x in fs range
• Example
• Show that f(x) x3 and g(x) x1/3 are inverses
  of each other
• g(f(x)) (x3)1/3 x for all real numbers (since
  fs domain is all real numbers)
• f(g(x)) (x1/3)3 x for all real numbers (since
  gs domain is all real numbers)
• To find an inverse of a function do the
  following
• Write the function as y f(x)
• Solve the equation for x in terms of y (if
  possible)
• Interchange x and y, the result is the function
  f-1(x) y

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Examples

• Find the inverse for the following
• f(x) 3x 2
• Rewrite as y y 3x 2
• Solve for x 3x y 2 ? x (y 2)
  / 3
• Exchange x for y y (x 2) / 3
• Replace y with f(x) f(x) (x 2) / 3
• So the inverse of 3x 2 is (x 2) / 3
• We check our work by confirming that f(f-1(x))
  x, 3(x2)/3) 2 x 2 2 x
• f(x) (x5 3) / 2
• y (x5 3) / 2
• x5 2y 3 or x (2y 3)1/5
• y (2x 3)1/5
• So the inverse is f(x) (2x 3)1/5
• We check our work f(f-1(x)) (((2x3)1/5)5 3)
  / 2 (2x 3 3)/2 x

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Example

• Here we examine the graph of an inverse
• Find the inverse of f(x) square root (x 2)
• Graph both the function and its inverse
• f(x) (x 2)1/2 ? y (x 2)1/2 ? y2 (x 2)
  for y gt 0 ? x y2 2, swap x and y giving y
  x2 2 for x gt 0
• f(x) x2 2 for x gt 0
• The graph of f is below to the left and f below
  to the right
• notice that f and f are mirror images about f(x)
  x

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8.1 Systems of Equations

• Many problems consist of multiple equations that
  each have more than 1 variable of interest
• All of our techniques to this point have allowed
  us to solve 1 equation for 1 variable in terms of
  another
• A group of equations is called a system of
  equations and we want to solve all equations for
  all variables
• this requires new techniques which we cover here
• Example
• Gas station sells regular at 2.20 per gallon and
  premium at 3.00 per gallon. In one day the gas
  station has made 680 on 280 total gallons. How
  much of each was sold?
• Let x gallons of regular and y gallons of
  premium, we have two equations
• 2.20 x 3.00y 680
• x y 280
• How do we solve for x and y?
• Does x 200 and y 80 provide a solution? yes
• 2.20 200 3.00 80 680
• 200 80 280

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Substitution Method

• We can solve a system of equations by using one
  of the equations to solve for one variable and
  then replace the variable in the other equation
  with this substitution
• In our previous example, we use the second
  equation x y 280 and solve for y 280 x
• Now we replace y in the first equation with 280
  x giving us 2.20 x 3.00(280 x) 680 and
  solve for x ? 2.20x 3.00 x 840 680 ? 160
  .8x or x 160 / .8 200
• Now we use x 200 in our substitution above to
  find y 280 x 280 200 80, so x 200, y
  80
• The strategy
• solve for one variable in terms of the other in
  one equation
• replace the variable in the other equation with
  our substitution
• now the other equation has only one variable,
  solve for it
• use this variable to determine the value of the
  other variable

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Examples

• 2x y 1 3x 4y 14
• From the first equation, solve for y 1 2x
• Now replace y in the second as 3x 4(1 2x)
  14
• Solve for x gives 3x 4 8x 14 ? 5x -10, x
  -2
• Replace x -2 in our replacement giving y 1
  2(-2) 5
• So x -2, y 5 solves this system of equations
• x2 y2 100 3x y 10
• Solve for y in the second equation gives y 3x
  10
• Replace y in the first gives x2 (3x 10)2
  100 ? x2 9x2 60x 100 ? 10x2 60x 100
  100 ? 10x2 60x 0 ? 10x(x 6) 0, so x 0
  or x 6
• We have two answers for x, we use them both to
  find y, so y 30 10 -10, or y 36 10
  8
• So our answers are x 0, y -10 or x 6, y 8
• Note we can also plot the equations and see
  where they intersect as we did earlier in the
  semester, see figures 1-3 on pages 622-623

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Elimination Method

• While the substitution method is straightforward,
  there are some equations where an alternative
  method might be easier
• Given two equations, algebraically manipulate one
  so that the coefficient of one of the variables
  is the negative version of the coefficient of the
  variable in the other equation
• Add the two equations together the result is
  that one of the variables drops out because you
  have ax -ax 0
• Solve for the other variable
• Substitute the value of the other variable in one
  of the equations to solve for the first variable
• Example 3x 2y 14 x 2y 2
• Since we have a 2y and a -2y, all we have to do
  is add the two equations giving 3x x 2y 2y
  14 2 ? 4x 16
• So x 4
• Replace x 4 in either equation, we will use the
  first, gives 34 2y 14 ? 2y 2 or y 1
• Use 4 and 1 in the second equation to check our
  work ? 4 21 2

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Example

• 3x2 2y 26 5x2 7y 3
• We solve for y by multiplying the first equation
  by -5 and the second by 3 giving
• -15x2 10y -130 and 15x2 21y 9
• Adding them gives 11y -121, or y -11
• Now we solve for x
• 3x2 2(-11) 26 ? 3x2 48, or x2 16, so x
  /- 4
• So our answers are x 4, y -11 and x -4, y
  -11
• We have 1 solution for y and 2 for x, which is
  reasonable since we have x2
• You can see the graph of this solution in figure
  5 on page 625 which clearly illustrates that for
  both x values, y is the same (-11)

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Graphical Method

• An alternative approach to solving these
  problems, like solving equations and inequalities
  is to graph the equations
• find their intersections
• This approach might be easier if the functions
  are easy to graph, but if they are hard to graph,
  this approach is not encouraged
• Also, finding the precise values of the
  intersections may not be easy especially if you
  are not very precise in your graphs
• We wont be using this approach for this class,
  but see figures 6-8 on pages 626-627 as examples

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8.2 Systems of Linear Equations

• As we saw in the last section, we can have a
  system of two equations and two variables
• if both equations are linear equations, then we
  have a system of linear equations to solve
• we can use any of the approaches from last
  section (substitution, elimination, graphing)
• in general, systems of linear equations will have
• one solution two lines which intersect at one
  position, the (x, y) coordinate of which is the
  solution to the system of equations
• no solutions if the equations represent lines
  that are parallel to each other
• infinitely many solutions if the equations
  represent the same line

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Examples

• System with 1 solution
• 3x y 0 5x 2y 22
• Solve by eliminating y by multiplying the first
  equation by 2 giving 6x 2y 0 and add to 5x
  2y 22 giving 11x 22, x 2, y 6
• System with 0 solutions
• 8x 2y 5 -12x 3y 7
• multiply the first by 3 and the second by 2 gives
• 24x 6y 15 and -24x 6y 14
• adding them gives 0 29, since this is false, it
  means that no matter what x or y values we
  assign, we cannot find a solution
• a problem with 0 solutions is sometimes called
  inconsistent
• System with infinitely many solutions
• 3x 6y 12 4x 8y 16
• multiply first by -4 and second by 3
• -12x 24y -48 12x 24y 48
• if we add them, we get 0 0, so for any x, there
  will be a y that solves these equations, or an
  infinite number of solutions
• for example, with x 0, then y -2, if x 4
  then y 0, if x 12 then y 4
• a problem with infinitely many solutions is
  sometimes called dependent

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Modeling with Systems of Equations

• We can model problems using a system of equations
• in this case, we restrict our examination to
  systems of linear equations
• given the problem, identify the two variables of
  the problem
• set up the system of equations, in this case,
  there should be two independent pieces of
  information to form two equations
• solve the system of equations as covered in the
  last section
• interpret the results

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Example

• A woman rows a boat upstream 4 miles in 1 ½ hours
  and the return trip, traveling with the current
  takes 45 minutes
• how fast does she row relative to the water and
  at what speed is the current flowing?
• she is rowing against the current in one
  direction and with the current in the other
• lets call x her rowing speed in miles per hour
  and y the currents speed in miles per hour
• 3/2 x 3/2 y 4
• 3/4 x 3/4 y 4 ? multiply by 2 gives 3/2
  x 3/2 y 8
• Adding the two together gives 3 x 12, x 4
  miles per hour
• 3/2 4 3/2 y 4 ? 2 3/2 y, y 4/3
  miles per hour
• the woman is rowing at 4 miles per hour, the
  current is flowing at 4/3 or 1 1/3 miles per hour

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Example

• A vintner fortifies wine that contains 10
  alcohol by adding a 70 alcohol solution to it
• The resulting mixture consists of 16 alcohol and
  fills 1000 one-liter bottles
• How many liters of wine and how many liters of
  the alcoholic solution did he use?
• let x be the liters of wine and y be the liters
  of the solution
• x y 1000
• we know that the combined mixture filled 1000
  one-liter bottles
• .10 x .70 y .16 1000
• the 1000 liters consisted of 16 alcohol of which
  each liter of wine had 10 alcohol and each liter
  of the solution had 70 alcohol
• we simplify the second term by multiplying both
  sides by 100 to be
• x 7y 1600
• So we have x y 1000 x 7y 1600
• Multiply the first equation by -1 giving x y
  -1000 and add the two to get 6y 600
• So y 100 and this gives us x 900