Alright a pretest scribe! I guess I have some explaining to do.... about the questions on the pretest!

Question 1)

If f(x) = ln(sqr(x)) then the average rate of change on the interval [3,7] is... then 5 possible answers. Using Roberts method you punch the equation into the calculator and use that handy-dandy slope program and out pops the answer or you can do it manually by hand and come up with A) 0.106

Question 2)

Suppose that the number of bacteria in a certain organism grows over time and the number, N(t), of bacteria (measured in thousands) at any time is given by:

N(t) = t(2 + cos(t))^(4/3) + 3t

At approximately what time ,t, in the first ten days is the colony growing the fastest?

First and foremost I must remind you all that IN THIS CLASS CALCULATORS SHOULD BE IN RADIANS! It is truly *very* annoying to miss a mark because of your calculator...

Now entered into your calculator you turn on that oh so useful derivative-of-whatever-you-have-in-your-Y1-slot graph and you look for the peak! (Then use that maximum function to give you a value)

The answer is.... 5.18!

Question 3)

Limit as x approaches one of:

(2x^2 + x - 3)/(3x^2 -x -2)

is equal to.... ?.

Well first you try putting one in right away and find your dividing by 0!

Algebraic massage required...

These both factor readily and two of the terms reduce. Then 1 can be entered in and you end up with 5/5 or 1... which is the answer!

Question 4)

The graph of the second derivative of a function f is shown at the right. Which of the following statements are true (I realize that there is no graph there but there but there is on this slide)

I) The graph of f has an inflection point at x = -1.

II) The f graph is concave down on the interval (-1,3).

III) The graph of the derivative function f prime is increasing at x = 1.

Then you find which of those statements is true.

I) An inflection point on a second derivative graph would be a zero and there happens to be one where they say that there is an inflection point therefore it is true!

II) The second derivative measures concavity, since it is negative on the interval that they say then the statement is true!

III) The second derivative is a measure of the slope of the first derivative. Since the value is negative the slope of that point is negative so it is decreasing at that point making this statement false.

So the answer is I and II are true!

Question 5)

Suppose a function f is defined so that is has derivatives:

f'(x) = x^2(x-1) and f''(x) = x(3x-2)

Over what intervals is f both increasing and concave up?

Well for it to be increasing the first derivative must be positive. Finding the roots of that we find them to be x = 0 and 1. Then do a sign check by entering a value before, in between and after to find when it is positive. The process is repeated for the second derivative because where that is positive the graph is concave up. We find the overlapping positive areas to be when x > 1!

Question 6)

a) Condider the following table of data.

Estimate f'(5.2) as accurately as possible.

Well f' would be the slope at that point!

Since they are on equal intervals and there are values before and after 5.2 we are able to use the symmetric difference quotient to get a more accurate result. It is found by finding the slope of that point with the points before and after and averaging it which results in you getting -9/4.

b) Write and equation for the slope of the tangeant line at that point.

Well... you are given a point.... you are given a slope... so let's use point slope form shall we? and you end up with y - 8.8 = -2.25 (x - 5.2)

c) Use your answer in part (b) to approximate f'(5.26)

Since the two points are so close together we can just plug that value into the equation we made to get a fairy accurate answer... which ends up being 8.665.

d) What is the sign of f''(5.2)? Explain.

Well this one I got wrong. I drew a graph where this point really looked like a point of inflection and that's what I based my answer on. WRONG! Mr.K ends up saying something along the lines of "Using the algebra to prove a point is far better then the graphical approach and will be expected of you on the exam". Since f' is decreasing to the left and right of this point we can deduce that this point is also decreasing meaning the graph is concave down meaning that f'' is negative at that point! (Though I still think it could be a point of inflection, but I do see the logic in what was said).

Woo that was a workout for the fingers!

The next scribe will be DINO!

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