Functions and Algebra: Solutions

Here are some solutions to the Exercises to accompany the section Functions and Algebra.  For this section, you should assume the following function definitions:

  1. Let f be given by the following table:
    a f(a)
    -2
    -1
    0
    1
    2
    3
    4
    5    		 1
    -2
    4
    5
    -1
    0
    3
    2
  2. Let g(x) = 3x + 2.
  3. Let h be given by the following arrow diagram:

  4. Let p be the function which "cubes the input, then subtracts 3 from the result."
  5. Let q be given by the following set of ordered pairs: {(-2, 3), (-1, 1), (0, -2), (1, 0), (2, 4), (3, 5), (4, -1), (5, 2)}.
  6. Let r be given by the following plot:

  7. Let s(b) = (b - 2)/3.
  8. Let k be given by the following arrow diagram:

  9. Let t be the function which "takes the reciprocal of the input, then adds 1 to the result."
  10. Let w be given by the following table:
    x w(x)
    -2
    -1
    0
    1
    2
    3
    4
    5    		 -1
    0
    3
    2
    -2
    4
    5
    1

Evaluation and Composition

  1. Using the definitions given above, evaluate each of the following composite functions at the given input value:
    1. Solution
      By the definition of the composite function:

        .  

      Working inside-out, we first compute g(-1) = 3(-1) + 2 = -1 (using the definition of g and remembering to use parentheses).  Substituting this result back in the previous equation gives f(g(-1)) = f(-1) = -2 (by the definition of f), and so

    2. Solution
      As in part a), , but now f(-1) = -2, and so we compute g(-2) = 3(-2) + 2 = -4.  This gives   Notice that, as we said before, the order of composition matters, since is not the same as
    3. Solution
      As in part a), , but we use the definition of k to compute k(3) = -1, and the definition of h to compute h(-1) = 0, so that .
    4. Solution
      As before, , and the definition of s gives s(11) = (11 - 2)/3 = 3.  From the graph of r, it appears that r(3) = 1, so that .
    5. Solution
      As usual, , and the description of p says that p(2) = 23 - 3 = 5.  Since the definition of q includes the ordered pair (5, 2), q(5) = 2, so that .
    6. Solution
      Finally, , and the description of t says that t(1/2) = 1/(1/2) + 1 = 2 + 1 = 3 (again, remembering to use parentheses).  From the definition of w, w(3) = 4, so that .

    Back to Exercises.

  2. Compute the indicated description for each of the following composite functions:
    1. Give a verbal description of .
      Solution
      Since the composite function is the result of first performing p then performing t, we can describe this function by "first cube the input, subtract 3 from the result, then take the reciprocal of that, and add 1 to the result."  Note: As a formula, this would look like .
    2. Give a formula for .  Hint: First determine the formula for p.
      Solution
      As a formula, p(x) =  x3 - 3, so that, by substitution, .
    3. Give the table of values for .  Hint: Compute the value of the composite on each possible input.
      Solution
      If we go down the first column of w and compute the value of on each possible input (remembering to apply w first and f second), we obtain:
      x
      -2
      -1
      0
      1
      2
      3
      4
      5      		 -2
      4
      0
      -1
      1
      3
      2
      5
    4. Give an arrow diagram for .  Hint: Compute the value of the composite on each possible input, then draw a picture of the result.
      Solution
      Simply placing the arrow diagrams for h and k together  k first and h second) gives:

      Eliminating the middle column, we obtain:

    5. Give a formula for .
      Solution
      Substituting the formula for g into that of s gives s(g(x)) = s(3x + 2) = ((3x + 2) - 2)/3 = (3x + 2 - 2)/3 = (3x)/3 = x.  That is, the composite is the "trivial" function, id(x), which as no effect on the input!  In the next section, we will investigate this sort of "cancellation" effect in more detail. 

    Back to Exercises.

Solving and Inverse Functions

  1. Using the definitions given above, give all possible solutions to each of following equations:
    1. f(x) = 5.
      Solution
      Looking in the second column of f:
      a f(a)
      -2
      -1
      0
      1
      2
      3
      4
      5      		 1
      -2
      4
      5
      -1
      0
      3
      2

       we see that 5 only appears as the output for the input of 1, in the first column.  Therefore, x = 1 is the only solution.

    2. h(x) = 2.
      Solution
      Looking in the second column of h:

      we see that 2 is pointed to by two arrows, from 1 and 3.  Therefore,  x = 1 or  x = 3 are the only solutions.
    3. p(x) = 24.
      Solution
      Thinking about the rule for p, we notice that 24 is 3 less than 27, which is 3 cubed.  Since there is only one cube root of 27, x = 3 is the only solution.
    4. r(x) = 4.
      Solution
      Looking in the plot of r, we see that there are two points on the graph at the same level as 4 on the vertical axis:

      One point is on the vertical axis, where x = 0, and the other appears to be above x = -2 on the horizontal axis.  Therefore,  x = 0 or  x = -2 are the only solutions.
    5. q(x) = -1.
      Solution
      Looking in the second entries of q, we see that -1 only appears in the ordered pair (4, -1), with a corresponding input of 4.  Therefore, x = 4 is the only solution.
    6. w(x) = 3.
      Solution
      Looking in the second column of w:
      x w(x)
      -2
      -1
      0
      1
      2
      3
       4
      5      		 -1
      0
      3
      2
      -2
      4
       5
      1
       
      we see that 3 only appears as the output for the input of 0, in the first column.  Therefore, x = 0 is the only solution.

    Back to Exercises.

  2. Using the definitions given above, compute the reverse in the indicated form of each of following functions:

    1. The reverse of f as a table.
      Solution
      To obtain the reverse relation, we switch columns:
      1
      -2
      4
      5
      -1
      0
      3
      2      		 -2
      -1
      0
      1
      2
      3
      4
      5
    2. The reverse of h as an arrow diagram.
      Solution
      To obtain the reverse relation, we reverse the arrows to get:

      or

    3. The reverse of p in words.
      Solution
      Since p first "cubes" and then "subtracts 3", the reverse would first "add 3" and then "take the cube root".
    4. The reverse of r as a plot.  Hint: First make a table of 4 - 6 points, reverse the table, then plot.
      Solution
      If we reverse the plot, switching the horizontal and vertical axes, the graph would be pointing to the right.  If we first make a table of points on the graph:
      -3
      -2
      -1
      0
      1
      2
      3      		 3
      4
      5
      4
      3
      2
      1
      we can calculate points on the reversed graph:
      3
      4
      5
      4
      3
      2
      1      		 -3
      -2
      -1
      0
      1
      2
      3
       and plot the results:

       

    5. The reverse of q as a table.
      Solution
      To obtain the reverse relation, we put the 2nd coordinates in the 1st column and the corresponding 1st coordinates in the 2nd column:
      3
      1
      -2
      0
      4
      5
      -1
      2      		 -2
      -1
      0
      1
      2
      3
      4
      5
    6. The reverse of w as an arrow diagram.
      Solution
      We can make a arrow diagram for w, and then reverse the arrows to get:

      or

    7. The reverse of t as a formula.
      Solution
      Since t first "takes the reciprocal" and then "adds 1", the reverse would first "subtract 1" and then "take the reciprocal".  That is because to get back the original input to a reciprocal, such as 1/2, we only need to take a reciprocal again; for example, 1/(1/2) = 2. 
        The formula which "subtracts 1" and then "takes the reciprocal" is 1/(x - 1) (simply follow the instructions on a typical input, x).

    Back to Exercises.

  3. Explain why each of the functions from the previous Exercise are or are not invertible.

    1. f.
      Solution
      From part a) of the previous exercise, we can see that the reverse relation of f is a function (i.e., only one output for each input), so f is invertible.
    2. h.
      Solution
      From part b) of the previous exercise, we can see that the reverse relation of h is not function, since an "input" of 0 has two possible outputs, -1 and 2; in terms of h itself, h(x) = 0 has two solutions, x = -1 and x = 2.  Again, this implies that h is not invertible.
    3. p.
      Solution
      From part c) of the previous exercise, we can see that the reverse relation of p is a function, since the inverse relation gives a clearly defined functional rule (i.e., only one output for each input), so p is invertible.
    4. r.
      Solution
      From part d) of the previous exercise, we can see that the reverse relation of r is not function; it fails the vertical line test, say for the vertical line crossing the horizontal axis at 4:

      since it intersects the graph at points with vertical coordinates 0 and -2.  Equivalently, the graph of r itself fails the horizontal line test:

      so that r is not invertible.
    5. q.
      Solution
      From part e) of the previous exercise, we can see that the reverse relation of q is a function (i.e., only one output for each input), so q is invertible.
    6. w.
      Solution
      From part f) of the previous exercise, we can see that the reverse relation of w is a function, since there is only one arrow from each input; alternatively, w(x) = y has one solution for each output, y, so w is invertible.
    7. t.
      Solution
      From part g) of the previous exercise, we can see that the reverse relation of t is a function, since the inverse relation gives a clearly defined functional rule (i.e., only one output for each input), so t is invertible.

    Back to Exercises.

  4. This Exercise is designed to help you see why "division by 0" is "not allowed" in Algebra.

    1. Let s be the "multiply by 0" function.  Complete the following arrow diagram for s:

       

      Solution
      Since 0 times anything is 0, all arrows should lead to 0:

    2. How would you describe the "opposite" of s in words? 
      Solution
      Since the opposite of multiplication is division, the opposite of "multiplication by 0" would be "division by 0".
    3. As in the previous Exercise, explain why s is or is not invertible.
      Solution
      The reverse of s is not a function, since 0 has many possible "outputs".  This means that s is not invertible, i.e., "division by 0" is not a well-defined function. 

    Back to Exercises.

More on Inverse Functions

  1. Following the example in the text, and using the definitions given above, give all possible solutions to each of following equations. 
    1. 2p(x - 1) - 3 = 7.
      Solution
      Adding 3 to both sides and dividing by 2 gives p(x - 1) = 5.  Applying p -1 to both sides gives x - 1 = p -1(5) = (5 + 3)1/3 = 2, using the formula from part c) of a previous exercise.  Then, x - 1 + 1 = 2 + 1, so x = 3.
    2. (-1/2)f(3x - 1) + 5 = 3.
      Solution
      Subtracting 5 from both sides and multiplying by -2 gives f(3x - 1) = 4.  Applying f -1 to both sides gives 3x - 1 = f -1(4) = 0, using the table from part a) of a previous exercise.  Then, 3x - 1 + 1 = 0 + 1, so 3x = 1 and x = 1/3.
    3. 5t(2 - x) + 3 = 3.
      Solution
      Subtracting 3 from both sides and dividing by 5 gives t(2 - x) = 0.  Applying t -1 to both sides gives 2 - x = t -1(0) = 1/(0 - 1) = -1, using the formula from part g) of a previous exercise.  Then, 2 - x - 2 = -1 - 2 = -3, so -x = -3 and x = 3.
    4. -3h(x - 1) + 1 = -5.  Note: Since is not invertible, you will get more than one solution.
      Solution
      Subtracting 1 from both sides and dividing by -3 gives h(x - 1) = 2.  Since h takes the value 2 for inputs of 1 and 3, we have x - 1 = 1 or x - 1 = 3.  Then, adding 1 to both sides of both equations, either x = 2 or x = 4.

    Back to Exercises.

  2. For each of following functions, use the Theorem in text to compute its inverse.  Verify each answer by showing that it satisfies the cancellation equations. 
    1. If g(x) = 3x + 2, compute g -1.
      Solution
      By the Theorem, we first set y = g(x) = 3x + 2, and then solve for x.  Subtracting 2 from both sides and dividing by 3 gives (y - 2)/3 = (3x + 2 - 2)/3 = 3x/3 = x.  This gives the reverse relation x = g -1(y), so that g -1(y) = (y - 2)/3.  This corresponds with our intuition that the reverse of "multiply by 3, then add 2" is "subtract 2, then divide by 3".
    2. If p(x) = x3 - 3, compute p -1.
      Solution
      Again, we set y = p(x) = x3 - 3, and solve for x.  Adding 3 to both sides and taking a cube root gives (y + 3)1/3 = (x3 - 3 + 3)1/3 = (x3)1/3 = x3/3 = x.  This gives the reverse relation x = p -1(y) = (y + 3)1/3.  Again, this corresponds with our intuition that the reverse of "cube, then subtract 3" is "add 3, then take a cube root".
    3. If f(x) = 1/(x + 4), compute f -1.
      Solution
      Starting with y = f(x) = 1/(x + 4), we can take reciprocals of both sides to get 1/y = x + 4.  Subtracting 4 from both sides gives the reverse relation x = f -1(y) = 1/y - 4.  In words, the reverse of "add 4, then take a reciprocal" is "take a reciprocal, then subtract 4".  This means that the inverse of the reciprocal function is the reciprocal function (just as the inverse of multiplication by -1 is multiplication by -1).

    Back to Exercises.

  3. Solve each of following equations for x.  Leave your answer as a formula in y, involving an appropriate inverse function. 
    1. Solve 2p(x - 1) - 3 = y for x, in terms of p -1.
      Solution
      Adding 3 to both sides and dividing by 2 gives p(x - 1) = (y + 3)/2.  Applying p -1 to both sides gives x - 1 = p -1((y + 3)/2).  Then, x - 1 + 1 = p -1((y + 3)/2) + 1, so x = p -1((y + 3)/2) + 1.
    2. (-1/2)f(3x - 1) + 5 = y for x, in terms of f -1.
      Solution
      Subtracting 5 from both sides and multiplying by -2 gives f(3x - 1) = -2(y - 5).  Applying f -1 to both sides gives 3x - 1 = f -1(-2(y - 5)), and 3x - 1 + 1 = f -1(-2(y - 5)) + 1.  Then, 3x = f -1(-2(y - 5)) + 1 or x = (f -1(-2(y - 5)) + 1)/3.
    3. 5t(2 - x) + 3 = y for x, in terms of t -1.
      Solution
      Subtracting 3 from both sides and dividing by 5 gives t(2 - x) = (y - 3)/5.  Applying t -1 to both sides gives 2 - x = t -1((y - 3)/5).  Then, 2 - x - 2 = t -1((y - 3)/5) - 2, so -x = t -1((y - 3)/5) - 2 and x = 2 - t -1((y - 3)/5).

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