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Algebra & Trigonometry (6.1): Composite Functions

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Terms in this set (23)

  • What is a composite function?
    A composite function is formed by applying one function to the result of another, denoted as \(f \circ g\), meaning f composed with g.
  • How is the domain of a composite function \(f \circ g\) determined?
    The domain consists of all x in the domain of g for which g(x) is in the domain of f.
  • What does the notation \(f \circ g\) represent?
    It represents the composite function where g is applied first, then f is applied to the result of g.
  • If \(f(x) = 2x + 3\) and \(g(x) = x^2\), what is \((f \circ g)(x)\)?
    \((f \circ g)(x) = f(g(x)) = 2x^2 + 3\).
  • If \(f(x) = \sqrt{x}\) and \(g(x) = x - 1\), what is the domain of \(f \circ g\)?
    The domain is all x such that \(x - 1 \geq 0\), so \(x \geq 1\).
  • How do you find the domain of \(f \circ g\) when f and g have restrictions?
    Find the domain of g, then exclude values where g(x) is not in the domain of f.
  • What is the domain of \(g(x) = \frac{1}{x-3}\)?
    All real numbers except \(x = 3\).

  • Example (Finding Domain of a Composite Function)

    Suppose f(x) = 1/x + 2 and g(x) = 4/x + 1

    Find: (a) f . g (b) f . f

    Then find the domain of each composite function.

    Solution:

    (a) (f . g) (x) = x -1/2(x +1)

    Domain of f . g was found to be {x | x not equal to 1, -1}

    Why?

    Domain of g {x|x not equal to 1} ; exclude 1 from domain of f . g

    Then, look at the domain of f . g : {x | x not equal to -1}

    Therefore, the domain of f . g : {x | x not equal to 1, -1}

  • Suppose f(x) = 1/x + 2 and g(x) = 4/x + 1

    Find: (a) f . g (b) f . f

    Then find the domain of each composite function.

    (b) (f . f) (x) = x + 2/2x + 5

    Domain of f . g : {x | x not equal to -5/2, -2}

    Why?

    Find domain of g {x|x not equal to 2} ; exclude 2 from domain of f . g

    Then, look at the domain of f . g : {x | x not equal to -5/2}

    Therefore, the domain of f . g : {x | x not equal to -5/2, -2}

  • If \(f(x) = \frac{1}{x+4}\), what values must be excluded from the domain of \(f \circ g\)?
    Values of x where \(g(x) = -4\) must be excluded.
  • How do you verify if two composite functions are equal?
    Show that \(f \circ g = h \circ k\) for every x in their common domain.
  • What is a natural way to decompose a function like \(H(x) = (3x + 2)^{21}\) into composite functions?
    Let \(g(x) = 3x + 2\) and \(f(x) = x^{21}\), so \(H = f \circ g\).
  • What is the domain of \(f(x) = \frac{1}{x-1}\)?
    All real numbers except \(x = 1\).
  • How do you find the domain of \(f \circ g\) if \(g(x) = \frac{1}{x-2}\) and \(f(x) = \sqrt{x+3}\)?
    Exclude \(x=2\) from g's domain and solve \(\frac{1}{x-2} + 3 \geq 0\) for x.
  • What must be excluded from the domain of \(f \circ g\) if \(f(x) = \sqrt{x}\) and \(g(x) = \frac{1}{x-1}\)?
    Exclude \(x=1\) and any x where \(g(x) < 0\).
  • What is the domain of \(g(x) = \sqrt{x-2}\)?
    All real numbers where \(x \geq 2\).
  • How do you find functions f and g such that \(H = f \circ g\) for a given H?
    Decompose H into two functions where g is applied first and f is applied to g's output.

  • Example 1 (#15)

    Find functions f and g so that f . g = H when H(x) = 1/x + 1

    H is the reciprocal of x + 1

    Split H (x) into f(x) = 1/x and g(x) = x + 1

    f . g(x) = 1/x + 3

  • Example 2 (#15)

    Find functions f and g so that f . g = H when H(x) =(x^2 + 1)^50

    H is the binomial (x^2 + 1) ^50

    Split H (x) into f(x) = x^50 and g(x) = x^2 + 1

  • What is the domain of \(f(x) = \frac{1}{x}\)?
    All real numbers except \(x = 0\).
  • If \(f(x) = x^2 + 1\) and \(g(x) = 3x - 4\), what is \((g \circ f)(x)\)?
    \(g(f(x)) = 3(x^2 + 1) - 4 = 3x^2 - 1\).
  • What is the range of a composite function?
    The set of all possible outputs of \(f(g(x))\) after applying both functions.
  • Why must you check the domain of both f and g when finding the domain of \(f \circ g\)?
    Because the input must be valid for g, and the output of g must be valid for f.