Question:

Step 1: Find the derivative to locate critical points. We compute the derivative of $f(x) = 2x^3 - 15x^2 + 36x + 1$: $$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$$ Setting $f'(x) = 0$, we find critical points at $x = 2$ and $x = 3$. Step 2: Evaluate the function at critical points and endpoints. We need to evaluate $f$ at the endpoints and critical points within $[0,3]$: $$f(0) = 2(0)^3 - 15(0)^2 + 36(0) + 1 = 1$$ $$f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$$ $$f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$$ Step 3: Analyze the monotonicity of $f$ on $[0,3]$. From $f'(x) = 6(x-2)(x-3)$: - For $x \in [0,2)$: both factors $(x-2)$ and $(x-3)$ are negative, so $f'(x) > 0$ and $f$ is increasing. - For $x \in (2,3)$: $(x-2) > 0$ and $(x-3) < 0$, so $f'(x) < 0$ and $f$ is decreasing. Therefore, $f$ increases on $[0,2]$ and decreases on $[2,3]$. Step 4: Determine if $f$ is one-one. Since $f$ increases on $[0,2]$ and then decreases on $[2,3]$, the function is not monotonic on the entire domain $[0,3]$. This means different values of $x$ can map to the same value of $y$. For example, there exist distinct points in $[0,2)$ and $(2,3]$ that map to the same output value. Therefore, $f$ is **not one-one**. Step 5: Determine if $f$ is onto. The maximum value of $f$ on $[0,3]$ is $f(2) = 29$, and the minimum value is $\min(f(0), f(3)) = \min(1, 28) = 1$. Since $f$ is continuous and achieves all values between its minimum and maximum, the range of $f$ is $[1, 29]$, which equals the codomain. Therefore, $f$ is **onto**. **Conclusion:** The function $f$ is onto but not one-one. The answer is **Option 3**.