{"id":911,"date":"2015-11-12T18:37:58","date_gmt":"2015-11-12T18:37:58","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=911"},"modified":"2017-03-31T20:18:47","modified_gmt":"2017-03-31T20:18:47","slug":"find-the-domain-of-a-composite-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/find-the-domain-of-a-composite-function\/","title":{"raw":"Find the domain of a composite function","rendered":"Find the domain of a composite function"},"content":{"raw":"

As we discussed previously, the domain of a composite function<\/strong> such as [latex]f\\circ g[\/latex] is dependent on the domain of [latex]g[\/latex] and the domain of [latex]f[\/latex]. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as [latex]f\\circ g[\/latex]. Let us assume we know the domains of the functions [latex]f[\/latex] and [latex]g[\/latex] separately. If we write the composite function for an input [latex]x[\/latex] as [latex]f\\left(g\\left(x\\right)\\right)[\/latex], we can see right away that [latex]x[\/latex] must be a member of the domain of [latex]g[\/latex] in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that [latex]g\\left(x\\right)[\/latex] must be a member of the domain of [latex]f[\/latex], otherwise the second function evaluation in [latex]f\\left(g\\left(x\\right)\\right)[\/latex] cannot be completed, and the expression is still undefined. Thus the domain of [latex]f\\circ g[\/latex] consists of only those inputs in the domain of [latex]g[\/latex] that produce outputs from [latex]g[\/latex] belonging to the domain of [latex]f[\/latex]. Note that the domain of [latex]f[\/latex] composed with [latex]g[\/latex] is the set of all [latex]x[\/latex] such that [latex]x[\/latex] is in the domain of [latex]g[\/latex] and [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f[\/latex].<\/p>\r\n\r\n

\r\n

A General Note: Domain of a Composite Function<\/h3>\r\n

The domain of a composite function [latex]f\\left(g\\left(x\\right)\\right)[\/latex] is the set of those inputs [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

How To: Given a function composition [latex]f\\left(g\\left(x\\right)\\right)[\/latex], determine its domain.<\/h3>\r\n
  1. Find the domain of g.<\/li>\r\n\t
  2. Find the domain of f.<\/li>\r\n\t
  3. Find those inputs,\u00a0x,\u00a0in the domain of g\u00a0for which g(x)\u00a0is in the domain of f. That is, exclude those inputs, x, from the domain of g\u00a0for which g(x)\u00a0is not in the domain of f. The resulting set is the domain of [latex]f\\circ g[\/latex].<\/li>\r\n<\/ol><\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 8: Finding the Domain of a Composite Function<\/h3>\r\n

    Find the domain of<\/p>\r\n\r\n

    [latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\frac{5}{x - 1}\\text{ and }g\\left(x\\right)=\\frac{4}{3x - 2}[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\n

    The domain of [latex]g\\left(x\\right)[\/latex] consists of all real numbers except [latex]x=\\frac{2}{3}[\/latex], since that input value would cause us to divide by 0. Likewise, the domain of [latex]f[\/latex] consists of all real numbers except 1. So we need to exclude from the domain of [latex]g\\left(x\\right)[\/latex] that value of [latex]x[\/latex] for which [latex]g\\left(x\\right)=1[\/latex].<\/p>\r\n\r\n

    [latex]\\begin{cases}\\frac{4}{3x - 2}=1\\hfill \\\\ 4=3x - 2\\hfill \\\\ 6=3x\\hfill \\\\ x=2\\hfill \\end{cases}[\/latex]<\/div>\r\n

    So the domain of [latex]f\\circ g[\/latex] is the set of all real numbers except [latex]\\frac{2}{3}[\/latex] and [latex]2[\/latex]. This means that<\/p>\r\n\r\n

    [latex]x\\ne \\frac{2}{3}\\text{or}x\\ne 2[\/latex]<\/div>\r\n

    We can write this in interval notation as<\/p>\r\n\r\n

    [latex]\\left(-\\infty ,\\frac{2}{3}\\right)\\cup \\left(\\frac{2}{3},2\\right)\\cup \\left(2,\\infty \\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 9: Finding the Domain of a Composite Function Involving Radicals<\/h3>\r\n

    Find the domain of<\/p>\r\n\r\n

    [latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\sqrt{x+2}\\text{ and }g\\left(x\\right)=\\sqrt{3-x}[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\n

    Because we cannot take the square root of a negative number, the domain of [latex]g[\/latex] is [latex]\\left(-\\infty ,3\\right][\/latex]. Now we check the domain of the composite function<\/p>\r\n\r\n

    [latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{3-x+2}\\text{ or}\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{5-x}[\/latex]<\/div>\r\n

    The domain of this function is [latex]\\left(-\\infty ,5\\right][\/latex]. To find the domain of [latex]f\\circ g[\/latex], we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since [latex]\\left(-\\infty ,3\\right][\/latex] is a proper subset of the domain of [latex]f\\circ g[\/latex]. This means the domain of [latex]f\\circ g[\/latex] is the same as the domain of [latex]g[\/latex], namely, [latex]\\left(-\\infty ,3\\right][\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Analysis of the Solution<\/h3>\r\n

    This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of [latex]f\\circ g[\/latex] can contain values that are not in the domain of [latex]f[\/latex], though they must be in the domain of [latex]g[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n
    \r\n
    \r\n

    Try It 6<\/h3>\r\n

    Find the domain of<\/p>\r\n\r\n

    [latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\frac{1}{x - 2}\\text{ and }g\\left(x\\right)=\\sqrt{x+4}[\/latex]<\/div>\r\n
    Solution<\/a><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    <\/div>","rendered":"

    As we discussed previously, the domain of a composite function<\/strong> such as [latex]f\\circ g[\/latex] is dependent on the domain of [latex]g[\/latex] and the domain of [latex]f[\/latex]. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as [latex]f\\circ g[\/latex]. Let us assume we know the domains of the functions [latex]f[\/latex] and [latex]g[\/latex] separately. If we write the composite function for an input [latex]x[\/latex] as [latex]f\\left(g\\left(x\\right)\\right)[\/latex], we can see right away that [latex]x[\/latex] must be a member of the domain of [latex]g[\/latex] in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that [latex]g\\left(x\\right)[\/latex] must be a member of the domain of [latex]f[\/latex], otherwise the second function evaluation in [latex]f\\left(g\\left(x\\right)\\right)[\/latex] cannot be completed, and the expression is still undefined. Thus the domain of [latex]f\\circ g[\/latex] consists of only those inputs in the domain of [latex]g[\/latex] that produce outputs from [latex]g[\/latex] belonging to the domain of [latex]f[\/latex]. Note that the domain of [latex]f[\/latex] composed with [latex]g[\/latex] is the set of all [latex]x[\/latex] such that [latex]x[\/latex] is in the domain of [latex]g[\/latex] and [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f[\/latex].<\/p>\n

    \n

    A General Note: Domain of a Composite Function<\/h3>\n

    The domain of a composite function [latex]f\\left(g\\left(x\\right)\\right)[\/latex] is the set of those inputs [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f[\/latex].<\/p>\n<\/div>\n

    \n

    How To: Given a function composition [latex]f\\left(g\\left(x\\right)\\right)[\/latex], determine its domain.<\/h3>\n
      \n
    1. Find the domain of g.<\/li>\n
    2. Find the domain of f.<\/li>\n
    3. Find those inputs,\u00a0x,\u00a0in the domain of g\u00a0for which g(x)\u00a0is in the domain of f. That is, exclude those inputs, x, from the domain of g\u00a0for which g(x)\u00a0is not in the domain of f. The resulting set is the domain of [latex]f\\circ g[\/latex].<\/li>\n<\/ol>\n<\/div>\n
      \n
      \n
      \n

      Example 8: Finding the Domain of a Composite Function<\/h3>\n

      Find the domain of<\/p>\n

      [latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\frac{5}{x - 1}\\text{ and }g\\left(x\\right)=\\frac{4}{3x - 2}[\/latex]<\/div>\n<\/div>\n
      \n

      Solution<\/h3>\n

      The domain of [latex]g\\left(x\\right)[\/latex] consists of all real numbers except [latex]x=\\frac{2}{3}[\/latex], since that input value would cause us to divide by 0. Likewise, the domain of [latex]f[\/latex] consists of all real numbers except 1. So we need to exclude from the domain of [latex]g\\left(x\\right)[\/latex] that value of [latex]x[\/latex] for which [latex]g\\left(x\\right)=1[\/latex].<\/p>\n

      [latex]\\begin{cases}\\frac{4}{3x - 2}=1\\hfill \\\\ 4=3x - 2\\hfill \\\\ 6=3x\\hfill \\\\ x=2\\hfill \\end{cases}[\/latex]<\/div>\n

      So the domain of [latex]f\\circ g[\/latex] is the set of all real numbers except [latex]\\frac{2}{3}[\/latex] and [latex]2[\/latex]. This means that<\/p>\n

      [latex]x\\ne \\frac{2}{3}\\text{or}x\\ne 2[\/latex]<\/div>\n

      We can write this in interval notation as<\/p>\n

      [latex]\\left(-\\infty ,\\frac{2}{3}\\right)\\cup \\left(\\frac{2}{3},2\\right)\\cup \\left(2,\\infty \\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n
      \n
      \n

      Example 9: Finding the Domain of a Composite Function Involving Radicals<\/h3>\n

      Find the domain of<\/p>\n

      [latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\sqrt{x+2}\\text{ and }g\\left(x\\right)=\\sqrt{3-x}[\/latex]<\/div>\n<\/div>\n
      \n

      Solution<\/h3>\n

      Because we cannot take the square root of a negative number, the domain of [latex]g[\/latex] is [latex]\\left(-\\infty ,3\\right][\/latex]. Now we check the domain of the composite function<\/p>\n

      [latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{3-x+2}\\text{ or}\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{5-x}[\/latex]<\/div>\n

      The domain of this function is [latex]\\left(-\\infty ,5\\right][\/latex]. To find the domain of [latex]f\\circ g[\/latex], we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since [latex]\\left(-\\infty ,3\\right][\/latex] is a proper subset of the domain of [latex]f\\circ g[\/latex]. This means the domain of [latex]f\\circ g[\/latex] is the same as the domain of [latex]g[\/latex], namely, [latex]\\left(-\\infty ,3\\right][\/latex].<\/p>\n<\/div>\n

      \n

      Analysis of the Solution<\/h3>\n

      This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of [latex]f\\circ g[\/latex] can contain values that are not in the domain of [latex]f[\/latex], though they must be in the domain of [latex]g[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n
      \n
      \n

      Try It 6<\/h3>\n

      Find the domain of<\/p>\n

      [latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\frac{1}{x - 2}\\text{ and }g\\left(x\\right)=\\sqrt{x+4}[\/latex]<\/div>\n
      Solution<\/a><\/div>\n<\/div>\n<\/div>\n<\/div>\n
      <\/section>\n<\/div>\n\n\t\t\t
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