More Operations with Functions: Solutions

Here are some solutions to the Exercises to accompany the section More Operations with Functions.  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 "squares the input, then subtracts 1 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 s(b) = (b - 2)/3.
  7. Let k be given by the following arrow diagram:

  8. Let t be the function which "takes the reciprocal of the input, then adds 1 to the result."
  9. 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

Arithmetic with Functions

  1. Using the definitions given above, evaluate each of the following functions at the given input value:
    1. (f + w)(4)
      Solution
      By the definition of the sum of functions, the value of the sum is the sum of the values, i.e., (f + w)(4) = f(4) + w(4).  Using the definitions for f and w, they evaluate at 4 to 3 and 5, respectively, so that (f + w)(4) = 3 + 5 = 8.
    2. (s - g)(2)
      Solution
      By the definition of the difference of functions, the value of the difference is the difference of the values, i.e., (s + g)(2) = s(2) + g(2).  Using the definitions for s and g, they evaluate at 2 to (2 - 2)/3 = 0 and 3·2 + 2 = 8, respectively, so that (s + g)(2) = 0 - 8 = -8.
    3. (h · k)(3)
      Solution
      By the definition of the product of functions, (h · k)(3) = (h(3))·(k(3)).  Using the definitions for h and k, they evaluate at 3 to 2 and -1, respectively, so that (h · k)(3) = 2·(-1) = -2.
    4. (p/q)(2)
      Solution
      By the definition of the quotient of functions, (p/q)(2) = (p(2))/(q(2)).  Using the definitions for p and q, they evaluate at 2 to 22 - 1 = 3 and 4, respectively, so that (p/q)(2) = 3/4 = 0.75.
    5. (3t)(5)
      Solution
      Since (3t)(5) = 3(t(5)), using the definition of t, we see that (3t)(5) = 3(1/5 + 1) = 3(1.2) = 3.6.

    Back to Exercises.

  2. Using the definitions given above, compute the indicated description for each of the following functions:
    1. Give an arrow diagram for h · k.  Hint: Compute the value of the product on each possible input, then draw a picture of the result.
      Solution
      Multiplying each pair of outputs, gives:

    2. Give the table of values for f + w.  Hint: Compute the value of the sum on each possible input.
      Solution
      Adding each pair of outputs, gives:
      x (f + w)(x)
      -2
      -1
      0
      1
      2
      3
      4
      5      		 1 + (-1) = 0
      (-2) + 0 = -2
      4 + 3 = 7
      5 + 2 = 7
      (-1) + (-2) = -3
      0 + 4 = 4
      3 + 5 = 8
      2 + 1 = 3
    3. Give a formula for p - g.  Hint: First determine the formula for p.
      Solution
      By the definition of the difference of functions, (p - g)(x) = p(x) - g(x).  Since the definition of p suggests that it is given by the formula p(x) = x2 - 1, (p - g)(x) = (x2 - 1) - (3x + 2) = x2 - 3x - 3.
    4. Give a formula for s/g.  
      Solution
      By the definition of the quotient of functions, (s/g)(x) = (s(x))/(g(x)).  Using the definitions for s and g, we have (s/g)(x) = ((x - 2)/3)/(3x + 2).
    5. Give a formula for 3t.  Hint: First determine the formula for t.
      Solution
      Now (3t)(x) = 3(t(x)), and since the definition of t suggests that it is given by the formula t(x) = 1/x + 1, (3t)(x) = 3(1/x + 1) = 3/x + 3.

    Back to Exercises.

Joining Functions

  1. Consider the following piecewise-defined function:

    Compute the following:

    1. f(-2)
      Solution
      Since -2 < -1, we are in the first case, so that f is given by the formula x2 and f(-2) = (-2)2 = 4.
    2. f(-1)
      Solution
      Since -1 is in the interval [-1, 1], we are in the second case, so that f is given by the formula 2x - 1 and f(-1) = 2(-1) - 1 = -3.
    3. f(0)
      Solution
      Since 0 is in the interval [-1, 1], we are still in the second case, so that f is given by the formula 2x - 1 and f(0) = 2(0) - 1 = -1.
    4. f(1)
      Solution
      Since 1 is in the interval [-1, 1], we are still in the second case, so that f(1) = 2(1) - 1 = 1.
    5. f(2)
      Solution
      Since 2 is neither less than -1, in the interval [-1, 1], nor greater than 3, f is not defined for this input (i.e., 2 is not in the domain of f), so f(2) is not defined.
    6. f(3)
      Solution
      Notice that f is not defined for an input of 3 either (since 3 < 3 is not true), so f(3) is not defined.
    7. f(4)
      Since 3 < 4, we are in the third case, where f is the square root function, so that f(4) = 2.

    Back to Exercises.

Decomposing Functions

  1. Express each of following functions in the indicated manner:
    1. If , determine formulas for functions g and k, so that , if h is the square root function.
      Solution
      We want , where h is the square root function.  This means that g is the function that is "inside" the square root (i.e., applied to the input before the square root), so we must have g(x) = -x + 3.  Likewise, k is the function that is "outside" the square root (i.e., applied to the result of the square root), so we must have k(x) = 4x - 5.  
    2. If f(x) = 4(-x + 3)2 - 5, determine formulas for functions g and k, so that , if h is the square function.
      Solution
      As before, , but this time h is the square function.  Looking "inside" the square, we see that again g(x) = -x + 3.  Moreover, the function that is applied to the result of the square is k(x) = 4x - 5.  Notice that g and k are the same functions as in part a), simply combined with a different function in the "middle".
    3. If , determine formulas for functions g, h, and k, so that f = k/g + h.
      Solution
      Since f(x) is the sum of (4x - 5)/(-x + 3) and x2, we see that the function k/g must be given by the formula (4x - 5)/(-x + 3), while h(x) = x2.  It is then clear that we should take k(x) = 4x - 5 and g(x) = -x + 3.  Notice that these are the same functions as in part b), just combined in a different way (i.e., via arithmetic operations, instead of composition).
    4. If f(x) = x2(2x + 1) - |3x - 4|, determine formulas for functions g, h, and k, so that f = k·g - h.
      Solution
      Since f(x) is the difference of x2(2x + 1) and |3x - 4|, we see that the function k·g must be given by the formula x2(2x + 1), while h(x) = |3x - 4|.  It is then clear that we should take k(x) = x2 and g(x) = 2x + 1.  

    Back to Exercises.

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