The function $$f:[0,3] \to [1,29]$$, defined by $$f(x) = 2x^3 - 15x^2 + 36x + 1$$, is :

First, calculate the derivative of $ f $ :

$$ f'(x) = 6x^2 - 30x + 36 $$

Which simplifies to :

$$ f'(x) = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3) $$

For the given domain $[0, 3]$, the function $ f(x) $ is both increasing and decreasing. Therefore, $ f(x) $ is many-to-one.

Setting $ f'(x) = 0 $ gives the critical points :

$$ x = 2, 3 $$

We now evaluate the function at these points within the domain to determine the range :

$$ f(0) = 1 $$

$$ f(2) = 29 $$

$$ f(3) = 28 $$

Thus, the range of $ f $ is $[1, 29]$, indicating that the function is onto.