Sorry it's so late. Kept forgetting and procrastinating. Yes, it was supposed to be me... but, I wasn't even participating during the class, because of APcalc in the corner of the room. But, I can at least explain the slides.

Slide 1, 2

Picture and the question.

Slide 3,4,5

Velocity is the change of distance over the change in time.

a. The average velocity, is the slope of the line, from the interval 0 to 4 of the given position function.

b. Velocity is zero when, the derivative of x(t) is equal to zero.

c. When the point is moving to the right (positive) meaning, the velocity is positive. (Derivative of x is positive)

d. Same thing as c, but moving left (negative)

e. When the derivative is equal to 3

f. Taking the derivative of the velocity function. Or the second derivative of the position function, called the acceleration function is equal to 3.

q. (Not sure why q, but okay) Draw what x(t) looks like. Red graph is the derivative of x(t)

Slide 6,7

The graph, and line test means, you are trying to find from the derivative, where there may be local minimums or maximums on the parent function.

So, when the derivative has values that crosses the x axis(zero) there is a local maximum or minimum. To determine whether it is a maximum or a minimum, on the left side of the zero, if it is a positive value then crosses the x-axis and then is negative, it is a local maximum. Vice versa, it is a local minimum.

Slide 8

The extreme value theorem, says, on a differentiated function, within a closed interval, there exists a maximum and minimum value on the parent function.

Slide 9 and 10 practice what we just learned.

That's it for that scribe. I'm onto Thursday's scribe now. Ciao.

## Monday, April 28, 2008

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