The function f: [0,3]→[1,29] defined by f(x) = 2x3 - 15x2 + 36x + 1 is

Concept: A function f: A → B is injective (one-to-one) when distinct inputs always produce distinct outputs, and is surjective (onto) when the values it produces cover the entire stated codomain B. By standard convention, interval notation like [0, 3] denotes every real number between 0 and 3 (a continuous interval), not just the integers 0, 1, 2, 3 — this convention is what allows the derivative and continuity argument below to apply.

Application: For f(x) = 2x3 − 15x2 + 36x + 1 on [0, 3]:

1. Differentiate: f'(x) = 6x2 − 30x + 36 = 6(x − 2)(x − 3).

2. Solve f'(x) = 0 within [0, 3]: the critical points are x = 2 and x = 3.

3. Check the sign of f'(x): it is positive on [0, 2) (f is increasing) and negative on (2, 3) (f is decreasing) — so f rises then falls, meaning it is not monotonic on [0, 3].

4. A non-monotonic function fails the horizontal-line test: since f increases up to x = 2 and then decreases, some output value is produced by two different inputs, so f is not injective.

5. Evaluate f at the boundary and critical points: f(0) = 1, f(2) = 29, f(3) = 28.

6. Because f is continuous, on [0, 2] it takes every value from 1 up to 29, and on [2, 3] it takes every value from 28 down to 29 — the union of these two ranges is exactly [1, 29].

7. [1, 29] is precisely the stated codomain, so every element of the codomain is attained, meaning f is surjective.

Cross-check: Pick a value in the overlap, say y = 28.5. Solving 2x3 − 15x2 + 36x + 1 = 28.5 gives one root in (0, 2), where f is rising toward 29, and a second root in (2, 3), where f is falling from 29 to 28. Two distinct inputs producing the same output independently confirms non-injectivity, while the endpoint values 1 and 29 confirm the full codomain is covered.

Result: So the function is surjective but not injective.