Relations and functions

High School Math 1 · Unit 7 · function-foundations · Teacher edition

function-foundations
Name ______________________________Date ______________
  1. 1.

    If f(x)=3x5f(x) = 3x - 5, find f(4)f(4).

    Answer: ______________

  2. 2.

    Let f(x)=x+42f(x) = \dfrac{x + 4}{2}. Find the value of xx for which f(x)=11f(x) = 11.

    Answer: ______________

  3. 3.

    Which relation is a function?

    1. (A)

      {(1,2), (2,3), (1,4)}\{(1, 2),\ (2, 3),\ (1, 4)\}

    2. (B)

      {(0,1), (0,2), (0,3)}\{(0, 1),\ (0, 2),\ (0, 3)\}

    3. (C)

      {(2,1), (2,3), (3,4)}\{(2, 1),\ (2, 3),\ (3, 4)\}

    4. (D)

      {(1,2), (2,2), (3,2)}\{(1, 2),\ (2, 2),\ (3, 2)\}

  4. 4.

    What is the domain of f(x)=x1f(x) = \sqrt{x - 1}, in interval notation?

    1. (A)

      (,)(-\infty, \infty)

    2. (B)

      [1,)[1, \infty)

    3. (C)

      (1,)(1, \infty)

    4. (D)

      [0,)[0, \infty)

  5. 5.

    The function C(m)=25+0.15mC(m) = 25 + 0.15m gives the cost in dollars of renting a van and driving it mm miles. For how many miles driven is the cost $61?

    Answer: ______________

Answer key — Relations and functions

  1. 1.
    7function-foundations-01

    f(4)=3(4)5=125=7f(4) = 3(4) - 5 = 12 - 5 = \mathbf{7}.

  2. 2.
    18function-foundations-02

    x+42=11x+4=22x=18\frac{x + 4}{2} = 11 \Rightarrow x + 4 = 22 \Rightarrow x = \mathbf{18}. Check: f(18)=222=11f(18) = \frac{22}{2} = 11. ✓

  3. 3.
    (D)

    {(1,2), (2,2), (3,2)}\{(1, 2),\ (2, 2),\ (3, 2)\}

    function-foundations-03

    In {(1,2),(2,2),(3,2)}\{(1, 2), (2, 2), (3, 2)\} the inputs 1,2,31, 2, 3 are all different, so each has exactly one output — it is a function even though every output is 22. Answer: {(1,2), (2,2), (3,2)}\mathbf{\{(1,2),\ (2,2),\ (3,2)\}}. (The first set sends 11 to both 22 and 44; the second sends 00 to three outputs; the third sends 22 to both 11 and 33.)

  4. 4.
    (B)

    [1,)[1, \infty)

    function-foundations-04

    x10x1x - 1 \ge 0 \Rightarrow x \ge 1, so the domain is [1,)\mathbf{[1, \infty)} with a square bracket because x=1x = 1 gives 0=0\sqrt{0} = 0. (All reals ignores the root; (1,)(1, \infty) wrongly excludes 11; [0,)[0, \infty) forgets the 1-1 inside.)

  5. 5.
    240function-foundations-05

    25+0.15m=610.15m=36m=360.15=24025 + 0.15m = 61 \Rightarrow 0.15m = 36 \Rightarrow m = \frac{36}{0.15} = \mathbf{240} miles. Check: 25+0.15(240)=25+36=6125 + 0.15(240) = 25 + 36 = 61. ✓