{"id":9217,"date":"2023-10-17T19:19:01","date_gmt":"2023-10-17T19:19:01","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&p=9217"},"modified":"2024-09-10T05:48:45","modified_gmt":"2024-09-10T05:48:45","slug":"function-basics-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/function-basics-learn-it-4\/","title":{"raw":"Function Basics: Learn It 4","rendered":"Function Basics: Learn It 4"},"content":{"raw":"

Finding Input and Output Values of a Function<\/h2>\r\n

When we know an input value and want to determine the corresponding output value for a function, we evaluate<\/em> the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.<\/p>\r\n

When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function\u2019s formula and solve<\/em> for the input. Solving can produce more than one solution because different input values can produce the same output value.<\/p>\r\n

Evaluation of Functions in Algebraic Forms<\/h3>\r\n

When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function [latex]f(x)=5\u22123x^2[\/latex] can be evaluated by squaring the input value, multiplying by [latex]3[\/latex], and then subtracting the product from [latex]5[\/latex].<\/p>\r\n

How to: Given the formula for a function, evaluate.<\/strong>\r\n
    \r\n\t
  1. Substitute the input variable in the formula with the value provided.<\/li>\r\n\t
  2. Calculate the result.<\/li>\r\n<\/ol>\r\n<\/section>\r\n
    \r\n

    When evaluating functions, it's handy to wrap the input variable in parentheses before making the substitution.<\/p>\r\n

    Ex. Given [latex]f(x)=x^2 - 8[\/latex], find [latex]f(-3)[\/latex]<\/p>\r\n

    [latex]\\begin{align}f(x)&=(x)^2 - 8 \\\\ &= (-3)^2 - 8 \\\\ &= 9 - 8 \\\\ &= 1\\end{align}[\/latex]<\/p>\r\n

    The value of the function [latex]f(x)=x^2 - 8[\/latex], at the input [latex]x=-3[\/latex], is [latex]1[\/latex].<\/p>\r\n<\/section>\r\n

    For the function, [latex]f\\left(x\\right)={x}^{2}+3x - 4[\/latex], evaluate each of the following.\r\n\r\n\r\n
      \r\n\t
    1. [latex]f\\left(2\\right)[\/latex]<\/li>\r\n\t
    2. [latex]f(a)[\/latex]<\/li>\r\n\t
    3. [latex]f(a+h)[\/latex]<\/li>\r\n\t
    4. [latex]\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/li>\r\n<\/ol>\r\n\r\n\r\n[reveal-answer q=\"645951\"]Show Solution[\/reveal-answer] [hidden-answer a=\"645951\"] Replace the [latex]x[\/latex]\u00a0in the function with each specified value.\r\n\r\n\r\n
        \r\n\t
      1. Because the input value is a number, 2, we can use algebra to simplify.\r\n

        <\/p>\r\n

        [latex]\\begin{align}f\\left(2\\right)&={2}^{2}+3\\left(2\\right)-4 \\\\ &=4+6 - 4 \\\\ &=6\\hfill \\end{align}[\/latex]<\/p>\r\n<\/li>\r\n\t

      2. In this case, the input value is a letter so we cannot simplify the answer any further.\r\n

        <\/p>\r\n

        [latex]f\\left(a\\right)={a}^{2}+3a - 4[\/latex]<\/p>\r\n<\/li>\r\n\t

      3. With an input value of [latex]a+h[\/latex], we must use the distributive property.\r\n

        <\/p>\r\n

        [latex]\\begin{align}f\\left(a+h\\right)&={\\left(a+h\\right)}^{2}+3\\left(a+h\\right)-4 \\\\[2mm] &={a}^{2}+2ah+{h}^{2}+3a+3h - 4 \\end{align}[\/latex]<\/p>\r\n<\/li>\r\n\t

      4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that:\r\n

        <\/p>\r\n

        [latex]f\\left(a+h\\right)={a}^{2}+2ah+{h}^{2}+3a+3h - 4[\/latex]<\/p>\r\n\r\n\r\nand we know that:\r\n\r\n\r\n

        [latex]f\\left(a\\right)={a}^{2}+3a - 4[\/latex]<\/p>\r\n\r\n\r\nNow we combine the results and simplify.\r\n\r\n\r\n

        [latex]\\begin{align}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}&=\\dfrac{\\left({a}^{2}+2ah+{h}^{2}+3a+3h - 4\\right)-\\left({a}^{2}+3a - 4\\right)}{h} \\\\[2mm] &=\\dfrac{2ah+{h}^{2}+3h}{h}\\\\[2mm] &=\\frac{h\\left(2a+h+3\\right)}{h}&&\\text{Factor out }h. \\\\[2mm] &=2a+h+3&&\\text{Simplify}.\\end{align}[\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n\r\n\r\n[\/hidden-answer]<\/section>\r\n

        Functions can be evaluated for negative values of [latex]x[\/latex], too. Keep in mind the rules for integer operations.<\/p>\r\n

        Given [latex]p(x)=2x^{2}+5[\/latex], find [latex]p(\u22123)[\/latex]. [reveal-answer q=\"489384\"]Show Solution[\/reveal-answer] [hidden-answer a=\"489384\"] Substitute [latex]-3[\/latex] in for x <\/i>in the function.\r\n\r\n\r\n

        [latex]p(\u22123)=2(\u22123)^{2}+5[\/latex]<\/p>\r\n\r\n\r\nSimplify the expression on the right side of the equation.\r\n\r\n\r\n

        [latex]\\begin{array}{lll}p(\u22123)=2(9)+5\\\\p(\u22123)=18+5\\\\p(\u22123)=23\\end{array}[\/latex]<\/p>\r\n\r\n\r\n[\/hidden-answer]<\/section>\r\n

        \r\n

        [ohm2_question hide_question_numbers=1]13509[\/ohm2_question]<\/p>\r\n<\/section>\r\n

        In addition to evaluating functions<\/strong> for a particular input, we can also solve functions<\/strong> for the input that creates a particular output.<\/p>\r\n

        How to: Solve a Function.<\/strong>\r\n
          \r\n\t
        1. Replace the output in the formula with the value provided.<\/li>\r\n\t
        2. Solve for the input variable that makes the statement true.<\/li>\r\n<\/ol>\r\n<\/section>\r\n
          Given the function [latex]h\\left(p\\right)={p}^{2}+2p[\/latex], solve for [latex]h\\left(p\\right)=3[\/latex]. [reveal-answer q=\"119909\"]Show Solution[\/reveal-answer] [hidden-answer a=\"119909\"]\r\n\r\n\r\n

          [latex]\\begin{align}&h\\left(p\\right)=3\\\\ &{p}^{2}+2p=3 &&\\text{Substitute the original function }h\\left(p\\right)={p}^{2}+2p. \\\\ &{p}^{2}+2p - 3=0 &&\\text{Subtract 3 from each side}. \\\\ &\\left(p+3\\text{)(}p - 1\\right)=0 &&\\text{Factor}. \\end{align}[\/latex]<\/p>\r\n\r\n\r\nIf [latex]\\left(p+3\\right)\\left(p - 1\\right)=0[\/latex], either [latex]\\left(p+3\\right)=0[\/latex] or [latex]\\left(p - 1\\right)=0[\/latex] (or both of them equal 0). We will set each factor equal to 0 and solve for [latex]p[\/latex] in each case.\r\n\r\n\r\n

          [latex]\\begin{align}&p+3=0, &&p=-3 \\\\ &p - 1=0, &&p=1\\hfill \\end{align}[\/latex]<\/p>\r\n\r\n\r\nThis gives us two solutions. The output [latex]h\\left(p\\right)=3[\/latex] when the input is either [latex]p=1[\/latex] or [latex]p=-3[\/latex].

          \"Graph<\/center>\r\n

           <\/p>\r\n\r\n\r\nWe can also verify by graphing, as seen above. The graph verifies that [latex]h\\left(1\\right)=h\\left(-3\\right)=3[\/latex] and [latex]h\\left(4\\right)=24[\/latex]. [\/hidden-answer]<\/section>\r\n

          \r\n

          [ohm2_question hide_question_numbers=1]13511[\/ohm2_question]<\/p>\r\n<\/section>","rendered":"

          Finding Input and Output Values of a Function<\/h2>\n

          When we know an input value and want to determine the corresponding output value for a function, we evaluate<\/em> the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.<\/p>\n

          When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function\u2019s formula and solve<\/em> for the input. Solving can produce more than one solution because different input values can produce the same output value.<\/p>\n

          Evaluation of Functions in Algebraic Forms<\/h3>\n

          When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function [latex]f(x)=5\u22123x^2[\/latex] can be evaluated by squaring the input value, multiplying by [latex]3[\/latex], and then subtracting the product from [latex]5[\/latex].<\/p>\n

          How to: Given the formula for a function, evaluate.<\/strong><\/p>\n
            \n
          1. Substitute the input variable in the formula with the value provided.<\/li>\n
          2. Calculate the result.<\/li>\n<\/ol>\n<\/section>\n
            \n

            When evaluating functions, it’s handy to wrap the input variable in parentheses before making the substitution.<\/p>\n

            Ex. Given [latex]f(x)=x^2 - 8[\/latex], find [latex]f(-3)[\/latex]<\/p>\n

            [latex]\\begin{align}f(x)&=(x)^2 - 8 \\\\ &= (-3)^2 - 8 \\\\ &= 9 - 8 \\\\ &= 1\\end{align}[\/latex]<\/p>\n

            The value of the function [latex]f(x)=x^2 - 8[\/latex], at the input [latex]x=-3[\/latex], is [latex]1[\/latex].<\/p>\n<\/section>\n

            For the function, [latex]f\\left(x\\right)={x}^{2}+3x - 4[\/latex], evaluate each of the following.<\/p>\n
              \n
            1. [latex]f\\left(2\\right)[\/latex]<\/li>\n
            2. [latex]f(a)[\/latex]<\/li>\n
            3. [latex]f(a+h)[\/latex]<\/li>\n
            4. [latex]\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/li>\n<\/ol>\n