HM

Domain Rules and Examples for Combined Functions (copy)

Domain Rules for Combining Functions

  • When we form a new function by combining two functions f and g via addition, subtraction, multiplication, or division, the domain of the resulting function is determined by the original domains of f and g. In other words, the new domain is a combination of the domains of f and g, and we must have something that works for both original functions. If a value of x makes either f or g undefined, the combined function is undefined at that x.
  • This principle holds for all four operations (addition, subtraction, multiplication, division).
  • For the quotient f ÷ g, there is an additional restriction: the denominator cannot be zero. In other words, g(x) cannot be zero for any x in the domain of the quotient. This restriction is the same restriction you would put on any division operation.
  • Therefore, the domain of f+g, f−g, and f×g is Dom(f) ∩ Dom(g). The domain of f÷g is Dom(f) ∩ Dom(g) with the extra requirement that g(x) ≠ 0 for all x in the domain.
  • In the example discussions, you’ll see these rules applied with explicit domains for sqrt-type and rational-type functions, and you’ll see how to determine where a denominator is zero or where a radical is defined.
  • A quick takeaway: always start with the domains of f and g, intersect them, and then apply any operation-specific restrictions (notably excluding x where the denominator is zero in division). If you simplify or cancel factors later, remember that the original domain before simplification might still exclude some x (holes can occur if a common factor cancels in a quotient).

Example 1: f(x) = \sqrt{1 - x}, g(x) = \dfrac{2x^2 + 1}{5x}

  • Goal: Find the domain for
    • a) f + g
    • b) f ÷ g
    • c) g ÷ f
  • Functions notation:
    • f(x) = \sqrt{1 - x}
    • g(x) = \dfrac{2x^2 + 1}{5x}

Part (a): Domain of f + g

  • Domain of f:
    • For a square root, the radicand must be nonnegative: 1 - x \ge 0 \quad\Rightarrow\quad x \le 1
    • Dom(f) = { x \in \mathbb{R} \mid x \le 1 } = (-\infty, 1]\n
  • Domain of g:
    • Denominator cannot be zero: 5x \neq 0 \quad\Rightarrow\quad x \neq 0
    • Dom(g) = { x \in \mathbb{R} \mid x \neq 0 } = (-\infty, 0) \cup (0, \infty)\n
  • Domain of f + g is Dom(f) ∩ Dom(g):
    • \text{Dom}(f+g) = (-\infty, 0) \cup (0, 1]
    • In set-builder form: { x \in \mathbb{R} \mid x \le 1 \text{ and } x \neq 0 }

Part (b): Domain of f ÷ g

  • Start with Dom(f) ∩ Dom(g) (same as part a), then apply the extra requirement that the divisor g(x) ≠ 0.
  • Check when g(x) = 0:
    • g(x) = \dfrac{2x^2 + 1}{5x} = 0 occurs when the numerator is zero (since denominator cannot be zero in the domain): 2x^2 + 1 = 0
    • This has no real solutions, so g(x) never equals 0 for real x in its domain.
  • Therefore, no additional exclusions arise beyond those in part (a).
  • Domain of f ÷ g is the same as in part (a): \text{Dom}(\tfrac{f}{g}) = (-\infty, 0) \cup (0, 1]

Part (c): Domain of g ÷ f

  • Start with Dom(g) ∩ Dom(f) as before: (-\infty, 0) \cup (0, 1]
  • Now require the denominator f(x) ≠ 0, i.e., f(x) = \sqrt{1 - x} \neq 0
    • Solve: \sqrt{1 - x} = 0 \Rightarrow 1 - x = 0 \Rightarrow x = 1
  • Exclude x = 1 from the domain:
    • So the domain becomes (-\infty, 0) \cup (0, 1)

Example 2 (DIY): f(x) = 5x^2 - 20, g(x) = (linear function)

  • Task: Find the domain of \tfrac{f}{g}(x)
  • Approach (as described):
    • Recognize that Dom(f) is all real numbers (a polynomial), and Dom(g) is all real numbers (a linear function).
    • Therefore, Dom(f) ∩ Dom(g) = \mathbb{R}.
    • The domain of the quotient requires that the denominator is not zero: require g(x) \neq 0\,.
  • Factoring f to see its structure (for later reference):
    • f(x) = 5x^2 - 20 = 5\left(x^2 - 4\right) = 5(x - 2)(x + 2)
    • The factorization highlights a difference-of-squares form and shows where f(x) is zero (at x = ±2). However, note that for the domain of the quotient, the zeros of f do not directly restrict the domain; rather, those zeros are relevant if you simplify the expression and cancel factors with g. The primary domain restriction for the quotient is still avoiding g(x) = 0.
  • Therefore, the domain of \tfrac{f}{g} is:
    • { x \in \mathbb{R} \mid g(x) \neq 0 }
  • If g is linear with a standard form g(x) = ax + b, then the zero of g is at x = -\frac{b}{a}, so the domain would be \mathbb{R} \setminus {-\frac{b}{a}}.
  • The note from the video emphasizes:
    • “Factoring out a GCF and then the difference of squares” is good practice when preparing for calculus and helps in recognizing structure, though it is not strictly required for the domain calculation of f/g.
    • The final simplified domain is the real line with the g-zero excluded (plus any domain restrictions inherited from f, if any, which in this example are none since f is a polynomial).

Takeaways and connections

  • Core idea: the domain of a combination of functions is governed by where each component is defined. Intersections reflect the need for both to be defined simultaneously.
  • Division introduces an extra constraint: the denominator cannot be zero; this is the primary cause for excluding x values beyond those required by the domains of f and g.
  • For radicals like sqrt, ensure the radicand is nonnegative. For rational expressions, ensure the denominator is nonzero.
  • When handling a quotient, be mindful of potential cancellations if you simplify; always refer to the original domain before simplification to avoid introducing extra holes.
  • Practical importance: knowledge of domain is essential before attempting to evaluate, differentiate, or integrate expressions built from f and g, and it informs where the function behaves well or may be undefined in real-world modeling.

Quick reference: key domains from the example

  • f(x) = \sqrt{1 - x}: \text{Dom}(f) = { x \in \mathbb{R} \mid 1 - x \ge 0 } = (-\infty, 1]
  • g(x) = \dfrac{2x^2 + 1}{5x}: \text{Dom}(g) = { x \in \mathbb{R} \mid x \neq 0 } = (-\infty, 0) \cup (0, \infty)
  • Domain results for Part (a): \text{Dom}(f+g) = (-\infty, 0) \cup (0, 1]
  • Domain results for Part (b): \text{Dom}(\tfrac{f}{g}) = (-\infty, 0) \cup (0, 1]
  • Domain results for Part (c): \text{Dom}(\tfrac{g}{f}) = (-\infty, 0) \cup (0, 1)

Additional notes (optional, for deeper understanding)

  • If you encounter a scenario where simplifying the expression cancels a factor, revisit the original domain to ensure no holes were introduced by cancellation. Holes in the graph correspond to points where the original denominator would have been zero or where the radicand would be negative, even if cancellation would remove the apparent restriction in the simplified form.
  • For the DIY example, the exact numeric domain of (f/g) depends on the explicit linear form of (g(x)). If (g(x) = ax + b) with (a \neq 0), then the domain is all real numbers except (x = -\tfrac{b}{a}). If you were given a specific linear function, you would substitute to identify the precise excluded value.