Well, you're in for a bug-tastic ride! To find the maximum velocity, we need to find the derivative of the position function, x(t), and then determine where it equals zero.
Taking the derivative of x(t), we get x'(t) = 3t² - 6t - 9. Now we just have to solve for when x'(t) = 0.
Let's set it equal to zero: 3t² - 6t - 9 = 0.
Now, to make things simpler, we can divide the equation by 3: t² - 2t - 3 = 0.
Ah, we now have a quadratic equation! To solve for t, we can either factor it or apply the quadratic formula. Let's go with the quadratic formula (X equals the opposite of b, so: t = (-(-2) ± sqrt((-2)^2 - 4(1)(-3))) / (2(1))). Crunching the numbers, we get t = 3 or t = -1. But since we're only considering the interval 0 <= t <= 2, we can throw the negative solution out the window.
So, the maximum velocity of our daring insect on the closed interval 0 <= t <= 2 occurs at t = 3. But hold on a second, there's a twist! The interval 0 <= t <= 2 doesn't contain t = 3! Uh-oh! So, the maximum velocity of our little critter on this interval is zero. It looks like this bug is taking a break and not going anywhere fast!
Remember, with a function this "buggy," you might encounter some unexpected twists and turns along the way. Keep on crawling, my friend!