BOB 5

Well, it's test time once again. I'm edging slightly towards holiday meltdown, so I don't think I'm in quite the right frame of mind to write this test, but that can't be helped. This is just a hard time of year for everyone, I think. I do feel that I understood most of the unit, though I find the optimization problems difficult, and I seem to need to see how to solve each specific type of problem, before I can solve it myself. It's generally the same sort of difficulty that I ran into with the word problems from last unit. Though I spent last unit focusing on those problems as they were my weakness, and there wasn't even one on the test! So I don't know what to do, there weren't any optimization problems or antiderivatives on the pretest. I don't know what I should focus on... And speaking of antiderivatives. I believe I have a good grasp of the idea behind them, but it is true that they are quite a lot harder than differentiating. I think they are the kind of thing that you just need to practice until the patterns become more familiar. And I hate it when stuff just doesn't exist.. imagine getting stuck on that for half an hour if your answer is impossible to find! And THANK YOU to Katrin for mentioning plus C in her bob, I was starting to forget about it. Well, I wish everyone luck and a nice relaxing vacation.

BOB #5

This chapter 5 unit on More Applications on the Derivative helped me understand the concept with derivatives a lot more. At first I never knew that you can use a number line to figure where there are max and mins and global mins and global max's. You can even use a number line to figure out where there are inflection points. I believe we use the second derivative to figure out inflection points. Besides the Line Test, I enjoy doing various problems like the classic optimizing problem. When I did a post on that I understood it more than if I were just to read it over. For me, making a post on work that I have learned helps me more. I also enjoy doing Antiderivatives. One thing that people might forget is to add (+) C. Remember to always add C! So test is tomorrow and I'm feel as if I need to review more of my notes because when I did the Practice Test, I didn't understand most of them. Though it’s a good thing we had group time to go through it, MR. K went over it in class and that Manny made a great post. It all helped!!!

Blogon Blog #5

Todays class was interesting! Pre-test today, yeaup.. scribe on it is down there. At first this unit was flowing like water to me, I got it all. I don't know happened... This pre-test was hard! At least for me. I got like 2 multiple choices right? And one part in the long answer? Ouch, it hurts. It might be because I haven't been doing alllll my homework, because I felt like it was enough. I guess not. I need to start cracking down full force on this calculus business. The optimization problems and related rate stuff are fun to me, because I know I can do it (most of them anyway). Most of my time was invested in those types of problems, but it wasn't on there. When I looked at the paper, I completely blanked as it looked completely new to me. It's an eye opener, cause the exam would probably be the same way! I think I just need to invest more time in more types of questions, and do allll of my homework, but that won't be till after the holidays (can't wait!). I just don't have anymore time. Well it's because I chose not to have any time by getting involved with things. Okay enough with my life story. I'm going to study study study for this test now, 3 hours already past just like that prior to doing my scribe earlier =(. GOODLUCK EVERYONE! And wish me luck , too... cause i need it =).

Scribe Post: Day 65

Well it's Manny and I'm your scribe for today! Todays class we did on our pre-test on our fifth unit on applications of derivatives. Here's the questions with the answer, and what I think the solutions are to get the answer.

Multiple Choice Part

1) The Graph of the function f(x)= 2x^5/3 - 5x^2/3 is increasing on which of the following intervals?I. 1 < x II. 0 < x < 1 III. x < 0The answer is I. and III. only.
- We get this by differentiating f into f'(x) = (10/3)x^2/3 - (10/3)x ^-1/3.
- Then we can factor it into f'(x) = (10/3)x ^-1/3(x-1) by taking out the lowest common factor to find its roots in order to make a number line
- We now know that there is a root at 0, and 1, so where ever it is positive, the function f is increasing.

2) Let f(x) = x⁵ - 3x² + 4. For how many inputs c between a=-2 and b=2 is it true that f(b) - f(a) / b - a = f'(c)
The answer is 2.
- First we'd find f(b) and f(a), which equals 24, and -40 respectively.
- Then plug it into the equation to get f'(c) to equal 16.
- Then differentiate f to get f'(x) = 5x⁴ - 6x.
- And since we're looking at the point f'(c), and we know what the point is, we get 16 = 5x⁴ - 6x.
- Then we'd just solve for the roots and we get the answer for how many we would have.
- I'd just punch in the equation the calculator and look at it over the [-2,2] interval and look at the zeros.

3) The table below gives some values of the derivative of a function g.


Based on this information it appears that on the interval covered by the table...
The answer is that g has a point of inflection.
- From the information, we know that g will be increasing everywhere, since it g' is positive throughout the interval.
- But we notice it g' starts to decrease after increasing, so at the highest point before decreasing there is a horizontal tangent line which means that there is a point of inflection because there is a change of sign from a positive to negative rate of change.

4) Suppose f is a continuous and differentiable function on the interval [0,1] and g(x) = f(3x). The table below gives some values of f

What is the approximate value of g'(0.1)?
The answer is approx. 3.84.
- For this we use the chain rule law.
- We know that g(x) = f(3x).
- So then we find g'(x) = f'(3x)(3) by the chain rule.
- After just plug in the values to get g'((.1)) = f'(3(.1))(3).
- Which turns into g'((.1)) = f'(.3)(3), then solve using the symmetric difference quotient.

5) If f(x) = ln(x) - k√(x) has a local minimum at x = 4 then the value of k is:
The answer is 1.
- If x = 4 is a minimum on the graph of f, then f'(4) = 0 must be true.
- So differiantiate f into f'(x) = 1/x - (1/2)k(x)^-1/2
- Then plug in the 4 f'(4) = 1/4 - (1/2)k(4)^-1/2
- Now solve for k.


Long Answer Part

1) Let f be a function given by f(x)= (3x - 2) / ( √(2x² + 1) )

a) Find the domain of f.
The answer is (-∞,∞)
When deciding domain, we need look only at the denominator. And in this function the denominator will never equal 0, meaning it's never undefined which means the function won't have vertical asymptotes. From that we know that the domain includes all the real numbers.

b) On the graph below, sketch the graph of f.


- We can sketch this by first looking where the function has roots. Roots occur when the function equals 0, or undefined. Look at both the numerator and denominator for roots. In this function the denominator can never equal 0 because of the square root function, which means there is no vertical asymptote. But in the numerator we notice that there is a root at 2/3.

- Then we can look at where we have y-intercepts, which occur when x=0. By plugging the 0 in for x into our function, it is easily located.

- Next, we can look at if theres any horizontal asymptote. And there is since the highest degree of power in the numerator is equal to the degree in the denominator. Horizontal asymptotes are discovered by the limits of infinity concepts. From this information we can just take the leading coefficient in the numerator which is 3, and divide it to the leading coefficient in the denominator which is √(2), so there is one at 3/√(2). And we can't forget about the -∞ side, to find that there is another horizontal asymptote at -3/√(2).

- Then we need to find out where the function is increasing and decreasing, and its concavity. For this we need the derivative. But the derivative was already given in part (d). So from finding the root to be -3/4, we make a number line to see that it is, decreasing to the left of it, and increasing to the right, which means there's a minimum at that point.

c) Write an equation for each horizontal asymptote of the graph of f.
The answer is lim x->∞ f(x)= 3/√(2) and lim x->-∞ f(x)= -3/√(2).
- Applying the infinity concept to our function. The -2 and 1, in the numerator and denominator respectively are insignificant to infinity, so we disregard it.
- So the denominator can be rewritten as √(2) * √(x²), which is √(2)*(x).

d) Find the range of f. [ Use f'(x) = 4x + 3 / (2x² + 1)^3/2 to justify your answer]
The answer is [ -17/4√(17/8), 3√(2) ).
- Since there's a minimum at x = -3/4, we can just plug in that value in our function to get the minimum output.
- Our maximum output is approaching 3/√(2) but never actually touching because of the asymptote there, found in part c).

Click here (part 1, part 2)to look at the paper with the correct circle answers and other multiple choices.

And that was it for the whole day. *Whew!* I'm a bit still unsure about parts of the long answer part, sooo if you're confused, I am too =). I tried my best to answer these problems in full detail as much to my understanding. Our group did pretty poorly on the test afterall. I'm scared for the test *gulp..* Anyway.. next scribe for Thursday will be Danny. Later days!

BOB

In this chapter, we have learned about the different applications of Derivatives. I really enjoyed this unit because I found it easy to understand. The only part that I had trouble with was the Optimization Problems. For me, it was very difficult to understand, especially the rate of travel. I have to clear this unit up before the test. I'll try to study hard on this part of the chapter. I hope everyone will do great on the test tomorrow!

BOB - Lindsay

This unit was a bit easier to understand than the previous units. Could it be that I'm getting better at this calculus business? I really do hope so. For the test, I think I should go over everything one more time because I usually miss SOMETHING. I'll somehow pack it all in this head of mine. I need some more practice on the optimization stuff and possibly antiderivatives but other than that, I think I'll be fine for the test. It's almost the winter break! Almost time to laze around and enjoy the luxury of doing nothing! I hope everyone does well on the upcoming test!

EDIT * nevermind I'm not ready. after that pre-test...boo it's 11:30 pm right now STUDY TIME.

BOB

MArk's BOB
I really enjoyed this unit. We did a lot of problem solving and group work for this unit. I found the problem solving the fun part in this unit. It was also fun to hear and learn with my fellow classmates. We had some disagreements and we had some clueless moments. We were there to learn as a group and we were there to help clarify anything that we found challenging. At this point in the course, i have realized that the way we think is different. At the beginning of the course, we were learning the basic idea about derivatives. Here we are now learning how to find anti-derivatives and we are learning how to solve optimization equations. In summary, i have not had much difficulty with this unit. I think it is understanding what steps to take to solve an equation/question. An example, solving word problems involving optimization.