The absolute minimum value of f(x)=3x^2-12x+5 on the interval [0,3] ()5 -4 7 3 -7

## Step 1: Find the derivative of the function<br />### To find the critical points, we need to take the derivative of the function $f(x) = 3x^2 - 12x + 5$. The derivative is $f'(x) = 6x - 12$.<br /><br />## Step 2: Find the critical points<br />### Set the derivative equal to zero and solve for $x$: $6x - 12 = 0 \Rightarrow x = 2$. This critical point lies within the interval $[0, 3]$.<br /><br />## Step 3: Evaluate the function at the critical point and endpoints<br />### Evaluate the function at the critical point $x=2$ and the endpoints $x=0$ and $x=3$:<br />$f(0) = 3(0)^2 - 12(0) + 5 = 5$<br />$f(2) = 3(2)^2 - 12(2) + 5 = 12 - 24 + 5 = -7$<br />$f(3) = 3(3)^2 - 12(3) + 5 = 27 - 36 + 5 = -4$<br /><br />## Step 4: Determine the absolute minimum<br />### Comparing the values $f(0)=5$, $f(2)=-7$, and $f(3)=-4$, the absolute minimum value is $-7$.