In the closed interval [0, 3], the minimum value of the function f given below is:

f(x) = 2x³ −9x² +12x

The correct option is 0.

To find the minimum value of the function $f\left(x\right)=2x^3-9x^2+12x$ in the closed interval [0, 3], we need to evaluate the function at its critical points within the interval and at the endpoints of the interval.

Step 1: Find the critical points of the function.
A critical point occurs where the first derivative $f\text{'}\left(x\right)$ is equal to zero or is undefined.
Let's find the first derivative of $f\left(x\right)$ with respect to $x$:
$f\text{'}\left(x\right)=\frac{d}{dx}(2x^3-9x^2+12x)$
$f\text{'}\left(x\right)=6x^2-18x+12$

Now, set the derivative to zero to find the critical points:
$6x^2-18x+12=0$
Divide the entire equation by 6 to simplify:
$x^2-3x+2=0$
Factor the quadratic equation:
$(x-1)(x-2)=0$
This gives the critical points:
$x=1$ and $x=2$

Both of these critical points, $x=1$ and $x=2$, lie within the given closed interval [0, 3].

Step 2: Evaluate the function at the critical points and the endpoints.
We need to calculate $f\left(x\right)$ at $x=0$, $x=1$, $x=2$, and $x=3$.

1. At the left endpoint, $x=0$:
$f\left(0\right)=2{\mathrm{(0)}}^3-9{\mathrm{(0)}}^2+12\left(0\right)=0$

2. At the critical point, $x=1$:
$f\left(1\right)=2{\mathrm{(1)}}^3-9{\mathrm{(1)}}^2+12\left(1\right)=2-9+12=5$

3. At the critical point, $x=2$:
$f\left(2\right)=2{\mathrm{(2)}}^3-9{\mathrm{(2)}}^2+12\left(2\right)=2\left(8\right)-9\left(4\right)+24=16-36+24=4$

4. At the right endpoint, $x=3$:
$f\left(3\right)=2{\mathrm{(3)}}^3-9{\mathrm{(3)}}^2+12\left(3\right)=2\left(27\right)-9\left(9\right)+36=54-81+36=9$

Step 3: Compare the values to find the absolute minimum.
Comparing the calculated values of $f\left(x\right)$:
- $f\left(0\right)=0$
- $f\left(1\right)=5$
- $f\left(2\right)=4$
- $f\left(3\right)=9$

The smallest of these values is 0, which occurs at the endpoint $x=0$. Therefore, the minimum value of the function on the closed interval [0, 3] is 0.