Given function

Check injectivity using derivative

Compute: $f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-2)(x-3).$

Now study the sign of $f'(x)$ on $[0,3]$:

• For $0\le x<2$, both $(x-2)$ and $(x-3)$ are negative, so $f'(x)>0.$
• For $2<x<3$, $(x-2)>0$ and $(x-3)<0$, so $f'(x)<0.$
• At $x=2,3$, $f'(x)=0$.

Hence $f$ is:

• increasing on $[0,2]$
• decreasing on $[2,3]$

Therefore, $f$ is not monotonic on $[0,3]$, so it cannot be one-one.

Also, we can directly verify by values: $f(1)=2-15+36+1=24,$ $f(3)=54-135+108+1=28.$ Since it increases then decreases, some values are repeated. So $f$ is not one-one.

Find the range to test surjectivity

Evaluate function at critical points and endpoints:

• At $x=0$: $f(0)=1.$
• At $x=2$: $f(2)=2(8)-15(4)+36(2)+1=16-60+72+1=29.$
• At $x=3$: $f(3)=28.$

Since $f$ is continuous on $[0,3]$, and:

• it increases from $1$ to $29$ on $[0,2]$,
• then decreases from $29$ to $28$ on $[2,3]$,

the overall range is $[1,29].$

The codomain is also $[1,29]$. Hence every value in $[1,29]$ is attained.

So $f$ is onto.

Conclusion

• Not one-one
• Onto

Therefore the correct option is: $\boxed{\text{B: onto but not one-one}}$

Comparison with stored correct answer

Stored correct answer = B.

Our derived answer is also B, so they agree.