The position at time t of an insect crawling along the x-axis is given by x(t) = t^3 - 3t^2 - 9t + 4. What is the maximum velocity on the closed interval 0 <= t <= 2?

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!

To find the maximum velocity on the closed interval 0 ≤ t ≤ 2, we need to find the derivative of the position function and then determine the critical points. The critical points occur where the derivative is equal to zero.

Therefore, let's find the derivative of the position function x(t):

x(t) = t^3 - 3t^2 - 9t + 4

To find the derivative, we differentiate each term with respect to t:

dx/dt = d(t^3)/dt - d(3t^2)/dt - d(9t)/dt + d(4)/dt
= 3t^2 - 6t - 9

Now, we need to find the critical points. Setting the derivative equal to zero:

3t^2 - 6t - 9 = 0

Factoring out a 3:

3(t^2 - 2t - 3) = 0

Using the quadratic formula with a = 1, b = -2, and c = -3:

t = (-(-2) ± sqrt((-2)^2 - 4(1)(-3))) / 2(1)
t = (2 ± sqrt(4 + 12)) / 2
t = (2 ± sqrt(16)) / 2
t = (2 ± 4) / 2

We have two solutions:

t = (2 + 4) / 2 = 6 / 2 = 3
t = (2 - 4) / 2 = -2 / 2 = -1

Since the interval is 0 ≤ t ≤ 2, we ignore the negative solution -1.

Therefore, the critical point on the interval 0 ≤ t ≤ 2 is t = 3.

Now, we need to evaluate the velocity at the critical point to determine if it is a maximum or minimum.

Substitute t = 3 into the derivative:

dx/dt = 3t^2 - 6t - 9
= 3(3)^2 - 6(3) - 9
= 27 - 18 - 9
= 0

The velocity at t = 3 is zero, which means it is neither maximum nor minimum. However, we also need to check the endpoints of the interval.

Substitute t = 0 into the derivative:

dx/dt = 3t^2 - 6t - 9
= 3(0)^2 - 6(0) - 9
= 0 - 0 - 9
= -9

Substitute t = 2 into the derivative:

dx/dt = 3t^2 - 6t - 9
= 3(2)^2 - 6(2) - 9
= 3(4) - 12 - 9
= 12 - 12 - 9
= -9

The velocity at both endpoints (t = 0 and t = 2) is -9.

Comparing the values, we can see that the maximum velocity on the closed interval 0 ≤ t ≤ 2 is -9.

To find the maximum velocity, we need to find the derivative of the position function, x(t), with respect to time, t.

The position function is given by x(t) = t^3 - 3t^2 - 9t + 4.

Taking the derivative of x(t) with respect to t, we get:

v(t) = dx(t)/dt = 3t^2 - 6t - 9.

Now, to find the maximum velocity, we need to check the critical points. Critical points occur when the derivative is equal to zero or is undefined.

Setting v(t) = 0, we have:

3t^2 - 6t - 9 = 0.

Factoring out the common factor of 3, we get:

3(t^2 - 2t - 3) = 0.

Using the quadratic formula, we can solve for t:

t = (-(-2) ± √((-2)^2 - 4(1)(-3))) / (2(1)).

Simplifying, we have:

t = (2 ± √(4 + 12)) / 2.

t = (2 ± √16) / 2.

t = (2 ± 4) / 2.

This gives us two possible values for t:

t = (2 + 4) / 2 = 3.

t = (2 - 4) / 2 = -1.

Since we are interested in the interval 0 <= t <= 2, the value t = -1 is not within this interval. Therefore, we can ignore it.

Next, we need to evaluate v(t) at the critical point t = 3 and at the endpoints of the interval 0 and 2.

v(0) = 3(0)^2 - 6(0) - 9 = -9.

v(2) = 3(2)^2 - 6(2) - 9 = 3.

v(3) = 3(3)^2 - 6(3) - 9 = 0.

From the evaluations, we see that v(0) = -9, v(2) = 3, and v(3) = 0.

Therefore, the maximum velocity on the closed interval 0 <= t <= 2 is 3.

find v(t), and then just take its derivative to find its maximum:

v(t) = 3t^2-6t-9
v'(t) = 6t-6
clearly, v'=0 at t=1
However, since v' is just a parabola, its vertex is a minimum, not a maximum.

That means that its maximum is either at t=0 or t=2. So, just compare v(0) and v(2).