The Second Derivative Test is failing?

Question

Define $f(x)$ as $2x^3-15x^2+36x+1$. Its domain is given to be $[0, 3]$ and range is $[1,29]$. Is it one-one? Is it onto?

I know that a monotonically increasing or decreasing function is one-one.
$$f'(x)= 6x^2-30x+36= 6(x-2)(x-3)$$ This is clearly positive for values of $x$ greater than three and negative for numbers in $(2,3)$. So the function is not one-one.
To find out whether the function is onto I found the maximum and minimum value of the function.
$$f''(x)= 12x-30$$ which is positive for $x=3$ and negative for $x=2$ where there would be a minimum and maximum respectively.
$f(2) = 29$ and $f(3)= 28$. So the value of $f(x)$ should range between $28$ and $29$. And the function is not onto. But clearly, $f(0)$ is equal to $1$.
I am not sure where I have gone wrong. I have checked my calculations multiple times. Is there a conceptual flaw in my solution? Please help me out.

Answer 1

The reason the test seems to fail is that the critical values you are considering, $x = 2,3$, are local extrema, not necessarily global extrema. Remember that the first/second derivative tests can only identify local extrema without additional information. To see that the function is onto in the domain $[0,3]$ note that $f$ is continuous, and that $f(0) = 1$ and $f(2) = 29$. By the Intermediate Value Theorem, then, $f$ is onto the range $[1,29]$.

Answer 2

Yes, the maximum of $f$ is attained at $2$. But you are wrong about the minimum. Since $f$ is strictly increasing on $[0,2]$ and strictly decreasing on $[2,3]$. The minimum is attained at $0$ or at $3$. And, since $f(0)=1$ and $f(3)=28$, the minimum is actually attained at $1$. So, yes, the range is $[1,29]$.

Answer 3

A local extremum can occur at three types of points:

1. Inner points where the derivative vanishes,
2. Points where the derivative is not defined,
3. Points on the boundary of the domain.

Here's the graph to make it easier to see what is happening. At the local extremum $(x,y)=(0,1)$ the derivative doesn't vanish, which is why you miss it as a possible extremum when you only look for points where the derivative vanishes.