Deciding Whether a Transformation Preserves the Domain of a Function
A pre-calculus class studies f(x) = √(x - 3) / (x - 7). A student performs a sequence of operations to 'simplify' the expression and obtain g(x) = 1 / (√(x - 3) + 4). The student argues that because g was derived from f by multiplying numerator and denominator by the conjugate √(x - 3) + 4, the two functions must be identical and therefore have the same domain. The teacher points out that f has restrictions the student has not accounted for. A second student claims the range of f is all real numbers because the numerator can be any real number. The class must determine the correct domain and range of f and decide whether f and g are the same function, and if not, how they differ.
1. Determine the domain of f and explain whether f and g are the same function.
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For f(x) = √(x - 3)/(x - 7), the square root requires x - 3 ≥ 0, so x ≥ 3. The denominator requires x - 7 ≠ 0, so x ≠ 7. Domain of f: [3, 7) ∪ (7, ∞). Now examine g(x) = 1/(√(x - 3) + 4). The square root again requires x ≥ 3. The denominator √(x - 3) + 4 is at least 4 for x ≥ 3, so it is never zero; no additional restriction. Domain of g: [3, ∞). The two functions differ at x = 7: f is undefined there, but g(7) = 1/(2 + 4) = 1/6. The algebraic manipulation that produced g is valid only where the multiplier √(x - 3) + 4 is nonzero, which it is, but the step also implicitly cancelled a factor of (x - 7) that was present in f. Multiplying by the conjugate is legitimate, but the resulting expression is equal to f only on f's domain. At x = 7 the original f has a removable discontinuity, while g is continuous. So f and g are not the same function; they agree on [3, 7) ∪ (7, ∞) but g extends continuously to x = 7.
2. Assess the second student's claim that the range of f is all real numbers.
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The claim is false. The numerator √(x - 3) is always nonnegative for x in the domain, so f(x) has the sign of the denominator x - 7. For 3 ≤ x < 7, the denominator is negative, so f(x) ≤ 0; at x = 3, f(3) = 0. For x > 7, the denominator is positive, so f(x) > 0. Therefore the range cannot include both signs at the same input, but more importantly the values are constrained. As x → 7⁻, the numerator is about 2 and the denominator approaches 0 from below, so f(x) → -∞. As x → 7⁺, f(x) → +∞. As x → ∞, f(x) ~ √x / x = 1/√x → 0⁺. So on (7, ∞), f decreases from +∞ toward 0, giving range (0, ∞). On [3, 7), f goes from 0 down to -∞, giving range (-∞, 0]. Combining: range is (-∞, 0] ∪ (0, ∞) = all real numbers? Careful: the value 0 is achieved at x = 3, and every positive value is achieved for x > 7, and every negative value for 3 < x < 7. So the range is indeed all real numbers, but the student's reasoning ('numerator can be any real number') is wrong because the numerator cannot be negative. The correct conclusion happens to be true, but for a different reason: the sign split across the vertical asymptote and the limits produce full coverage. The teacher should credit the conclusion only with correct justification.