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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The original domain of f(x) = sqrt(x - 3)/(x - 7) is [3,7) union (7,infinity): the radicand must be nonnegative and the denominator must be nonzero. The proposed g(x) = 1/(sqrt(x - 3) + 4) is defined on [3,infinity). They are not equal, even away from x = 7: at x = 3, f(3) = 0 while g(3) = 1/4. At x = 7 the numerator of f approaches 2, so f has a vertical asymptote, not a removable hole. Multiplying by a nonzero conjugate cannot justify the claimed expression; the student's asserted cancellation is an algebraic error.
2. Assess the second student's claim that the range of f is all real numbers.
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The conclusion that the range of f is all real numbers is correct, but the stated reason is wrong: sqrt(x - 3) is never negative. On [3,7), f is continuous, takes value 0 at x = 3, is negative thereafter and tends to negative infinity as x approaches 7 from below. The intermediate value theorem therefore supplies every nonpositive value. On (7,infinity), f is positive and continuous, tends to positive infinity near 7 and tends to 0 from above as x grows. It therefore supplies every positive value. The two branches together give the entire real line; the numerator alone does not establish the range.