The range of the function f(x) = x2/(1 + x2) is:

Concept: The range of f(x) = N(x)/D(x) is the set of y-values for which the equation y = f(x) has at least one real solution for x. Rearranging the equation to isolate x2 and requiring x2 ≥ 0 (since the square of a real number can never be negative) pins down exactly which y-values are attainable.

1. Set up and isolate x2:

   Let y = x2/(1 + x2) and rearrange: y(1 + x2) = x2, so y = x2(1 − y), giving x2 = y/(1 − y) (valid whenever y ≠ 1).

2. Apply the non-negativity constraint:

   Since x2 ≥ 0 for every real x, the fraction y/(1 − y) must also be ≥ 0.

3. Solve the inequality:

   y/(1 − y) ≥ 0 holds precisely when 0 ≤ y < 1 (y = 1 is excluded because the denominator becomes 0).

4. Check the lower boundary:

   At y = 0: x2 = 0 gives x = 0, a real solution, so y = 0 IS attained by the function.

5. Check the upper boundary:

   As y approaches 1 from below, x2 grows without bound, so y = 1 is approached but never actually reached by any real x.

Cross-check: This matches a direct check on the expression: at x = 0, f(0) = 0/(1 + 0) = 0, confirming the lower bound is attained; and since the numerator x2 is always exactly 1 less than the denominator 1 + x2, the quotient is always strictly less than 1, approaching it only as |x| grows without bound.

Result: Combining both checks, the function attains every value from 0 up to but not including 1 — the range is [0, 1).