{"id":1426,"date":"2024-05-23T03:59:55","date_gmt":"2024-05-23T03:59:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=1426"},"modified":"2025-08-13T16:01:19","modified_gmt":"2025-08-13T16:01:19","slug":"introduction-to-functions-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/introduction-to-functions-learn-it-4\/","title":{"raw":"Introduction to Functions: Learn It 4","rendered":"Introduction to Functions: Learn It 4"},"content":{"raw":"

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

Evaluation of Functions in Algebraic Forms<\/h3>\r\nWhen 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].\r\n\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>
    When evaluating functions, it's handy to wrap the input variable in parentheses before making the substitution.\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\nThe value of the function [latex]f(x)=x^2 - 8[\/latex], at the input [latex]x=-3[\/latex], is [latex]1[\/latex].\r\n\r\n<\/section>

    For the function, [latex]f\\left(x\\right)={x}^{2}+3x - 4[\/latex], evaluate each of the following.\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[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 \t
      1. Because the input value is a number, 2, we can use algebra to simplify.\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

        [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

        [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

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

        [latex]f\\left(a\\right)={a}^{2}+3a - 4[\/latex]<\/p>\r\nNow we combine the results and simplify.\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[\/hidden-answer]\r\n\r\n<\/section>Functions can be evaluated for negative values of [latex]x[\/latex], too. Keep in mind the rules for integer operations.\r\n\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 [latex]x[\/latex] in the function.\r\n

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

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

        [ohm2_question hide_question_numbers=1]13509[\/ohm2_question]<\/section>In addition to evaluating functions<\/strong> for a particular input, we can also solve functions<\/strong> for the input that creates a particular output.\r\n\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>
          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

          [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\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 [latex]0[\/latex]). We will set each factor equal to [latex]0[\/latex] and solve for [latex]p[\/latex] in each case.\r\n

          [latex]\\begin{align}&p+3=0, &&p=-3 \\\\ &p - 1=0, &&p=1\\hfill \\end{align}[\/latex]<\/p>\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].\r\n\r\n

          \r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Graph of a parabola with labeled points (-3, 3), (1, 3), and (4, 24).[\/caption]\r\n\r\n<\/center> \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]\r\n\r\n<\/section>
          [ohm2_question hide_question_numbers=1]13511[\/ohm2_question]<\/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
            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