October 26, 2006

Scribe Post: Day 32

Hi ya'll! This is JANN and I'll be your scribe for today.

Before going over what we did today, I just want to remind everyone about the test tomorrow. I hope everyone will do great! XD Anyways, here we go:

In the beginning of class, we went over the last 2 questions in the pre-test.

(5) Suppose a function "f" is defined so that it has derivatives:

Over what interval is the graph of "f" both increasing and concave up?


To identify what interval is "f" increasing and concave up, first find the roots of the two equations:

f'(x)= x^2(x-1)= (x^2+2)(x-1), x= 0,1
f"(x)= x(3x-2)= (x+0)(3x-2), x= 0,2/3

Then, use the number line method for each equation.

We can see that at the intervals x>1 and x>2/3, the answer is positive. Therefore, the graph is increasing due to f' and concave up due to f" since both are positive.
The answer is x>1. The number 1 is not included because the graph increases after x=1.

(6) Consider the following table of data for the function "f".

(a) Estimate f'(5.2) as acurately as possible.


In order to estimate f'(5.2), we use the Symmetric Difference Quotient method. Since the given is a table of values, we can get the slope to the left and to the right of x= 5.2 Then, we get the average of the 2 slopes.

(b) Write the equation for the tangent line at x= 5.2


To get the equation of the tangent line at x= 5.2, we could use the point-slope formula using the pt (5.2,8.8) and the answer to 6a.

y-y1= m(x-x1)
y-8.8= -9/4(x-5.2)

(c) Use your answer to part b to find approximately, f(5.26)


Using the formula found in part b, we could simply substitute f(5.26) as "y" and 5.56 as "x".

y-8.8= -9/4(x-5.2)
f(5.26)-8.8= -9/4(5.26-5.2)

(d) What is the sign of f"(5.2)? Explain your answer.


f"(5.2) is "negative" because the first derivative of the parent function is decreasing due to the slopes -2 and -2.25 Because the slopes are decreasing, the 2nd derivative at these points are negative.

After going over the last 2 questions, we were given another group problem. Here's the function of the problem.

I forgot how the questions were typed, so here are the answers anyways. XD

Solution: The graph of the function

The graph is constantly increasing from negative infinity up to x=3. As soon as it reaches x>3, the graph jumps up in a parabolic funcion. Therefore the graph is discontinuous and nonremovable. Its not a good idea to but grain from this company because if you buy more than 3 bags at $9, the next bags would be higher that the rate of the first 3 bags.

Finally, we were given 2 problems.


a) Is "f" continuous at x=3? NO
b) Determine f'(0)= 1
c) Determine f'(3)= does not exist


a) On what interval is the graph of "f" concave down? (-3,-2)
b) Where on the graph of "f" will there be a point of inflection? x=-2
c) If "f" is decreasing at x= -2.5, is "f" increasing or decreasing at x=-2.Explain Decreasing

Tip of the Day: Enter nDeriv(Y1,X,X) in your Y0 and leave it there until we throw away our calculator, I guess. =P If you put this on your calc, you dont have to re-enter this command in Y2 every time you need to find the derivative of the function in Y1.

I think thats all we did today. Sorry if my post is colorless(besides blue and black). I dont know why, but the tool bar on top is pretty... short? The color and font size menus are missing. 0.0 Anyways, good luck to ya'll tomorrow!! =D

Oh yah I almost forgot. Next Scribe for Monday is... Suzanne!

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