{"id":10708,"date":"2015-07-10T19:04:40","date_gmt":"2015-07-10T19:04:40","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=10708"},"modified":"2021-10-25T01:58:28","modified_gmt":"2021-10-25T01:58:28","slug":"evaluating-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/evaluating-functions\/","title":{"raw":"Evaluating functions","rendered":"Evaluating functions"},"content":{"raw":"

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\r\n

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Evaluation of Functions in Algebraic Forms<\/span><\/h2>\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\\left(x\\right)=5 - 3{x}^{2}[\/latex] can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.<\/p>\r\n\r\n

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How To: Given the formula for a function, evaluate.\r\n<\/strong><\/h3>\r\n
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  1. Replace the input variable in the formula with the value provided.<\/li>\r\n \t
  2. Calculate the result.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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    Example: Evaluating Functions<\/h3>\r\nGiven the function [latex]h\\left(p\\right)={p}^{2}+2p[\/latex], evaluate [latex]h\\left(4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"768180\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"768180\"]\r\n

    To evaluate [latex]h\\left(4\\right)[\/latex], we substitute the value 4 for the input variable [latex]p[\/latex] in the given function.<\/p>\r\n

    [latex]\\begin{align}h\\left(p\\right)&={p}^{2}+2p \\\\ h\\left(4\\right)&={\\left(4\\right)}^{2}+2\\left(4\\right) \\\\ &=16+8 \\\\ &=24 \\end{align}[\/latex]<\/p>\r\nTherefore, for an input of 4, we have an output of 24 or [latex]h(4)=24[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/Ehkzu5Uv7O0\r\n

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    Example: Evaluating Functions at Specific Values<\/h3>\r\nFor the function, [latex]f\\left(x\\right)={x}^{2}+3x - 4[\/latex], evaluate each of the following.\r\n
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    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]\r\n[hidden-answer a=\"645951\"]\r\n\r\nReplace the [latex]x[\/latex]\u00a0in the function with each specified value.\r\n
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      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]<\/div><\/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]<\/div><\/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]<\/div><\/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]<\/div>\r\nand we know that\r\n
        [latex]f\\left(a\\right)={a}^{2}+3a - 4[\/latex]<\/div>\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