**1. Suppose f(x) = √x.**

**a) Calculate the average rate of change between x = 9 and x = 9.1.**

f(9.1) - f(9) / 9.1 - 9.0 = average rate of change

**√9.1 - √9 / 0.1 = 0.1662**(Store the answer from your calc in A)

**b)**

**Calculate the average rate of change between x = 8.9 and x = 9.0.**

f(9) - f(8.9) / 9 - 8.9 = average rate of change

**√9 - √8.9 / 0.1 = 0.1671**(Store the answer from your calc in B)

**c) Use your answers above to approximate the instantaneous rate of change at x = 9.0.**

Take the answers from questions A and B:

**(A+B) / 2 = 0.16666**

NOTE: You must have your answers for

*A)*and

*B)*to four decimal places, you don't start seeing a change between the two answers until four decimal places. Marks normally aren't given out if your answers aren't to four decimal places. When calculating the answer to

*C)*in your calculator, you must have brackets around

*A + B*or the calculator will calculate

*B / 2*then

*+ A*into the solution.

**2. A ball is dropped from a height of 400 ft and falls towards the earth in a straight line. In**

*t*seconds the ball drops a distance of d = 16t² feet.**a) How many seconds after release does the ball hit the ground?**

The ball is dropped from a height of 400 ft, therefore the ball must travel 400 ft before it hits the ground:

400 = 16t²

400 / 16 = t²

**t = 5**

**b) What is the average velocity of the ball during the time it is falling?**

**0 - 400 / 5 - 0 = -80 ft/sec**

The reason why we take 400 and subtract it from 0 and not the other way around is because the ball is falling down not falling up, so the velocity is increasing in a negative manner.

**c) Estimate the instantaneous velocity of the ball when it hits the ground.**

**160 ft/sec**

There are many methods of finding this answer. You can graph it and keep zooming into that point and use the

**"SLOPE" program**, you can estimate by using the

**symetric difference quotient**and others. Unfortunately my mind is mashed right now and I can't recall any other from the top of my head. =S

**3. An object travels in such a way that its position at various points in time is given in the table:**

**a) Find the average velocity of the object between t = 1.3 and t = 1.9.**

5.3 - 5.1 / 1.9 - 1.3 =**0.2 / 0.6 = 1/3b) Estimate the instantaneous velocity at t = 1.9. Show work.**5.3 - 5.6 / 1.9 - 1.8 =

**-0.3 / 0.1 = -3**

While Mr. K was giving us the solutions to this quiz he talked about the exam for the course. Here's a rough outline of the exam:

- consists of two sections, multiple choice and long answer.
- each section has a non-calculator part and a calculator part.
- multiple choice has 28 questions non-calculator (50 min) and 17 calculator questions (50 min).
- long answer has 3 calculator questions (50 min) and 3 questions non-calculator (50 min).
- the questions and sections come in the order I just explained.
- the exam itself is a MARATHON 4 HOURS from 8 a.m. to 12 p.m.

Now onto the topic we talked about in class....or at least what I managed to write down of it.** **=S** "Limits"**

We started with a function **f(x) = 2x - 1**: **(Note: graph may be up tomorrow, it's too late for me honestly =S)**

We were given **lim x->2 f(x)** which is, **lim x->2 2x - 1**. If you were to try and find **x = 2** the answer would come up to be 3. But the **lim x->2** means you have a value very close to 2 your input and a value very close to 3 your output, but it never actually gets to the point **(2,3)**. **I'm not entirely sure if what I just explained made any sense I made a lot of point form notes and I'm trying to work from it I apologize now if any of my information is incorrect or has no value to it.**We were then given

**(x² - 1) / (x - 1)**and we graphed it on our calculators. When the graph was shown in a

**"Zoom Standard"**window it looked like just an ordinary straight line. But when you look at it in a

**"Zoom Decimal"**window, you see that the line in fact has a HOLE in it! The reason for this is because

**(x² - 1)**factors and reduces. The hole in the graph is at

**x = 1**because there is no value for that. When you look at the table there seems to be a limit at

**x = 1**. The

*y*values get closer and closer to 2 but never really gets there. Therefore

*x*can never equal 1 cause of the limit.

**lim x->1 (x²- 1) / (x - 1) = (x - 1)(x + 1) / (x - 1) = (x + 1)**

**This is were my notes get a little confusing =S.** Mr. K started talking about dividing by 0 and why it's not allowed. When you multiple 2 by 3 you get 6 and when you divide 6 by 3 you get 2, so when you multiple 2 by 0 you get 0 and when you divide 0 by 0 you get 2? So basically 0/0 can give you any number? We call this **"indeterminate".** If you divide 5 by 1 you get 5, if you divide 5 by 0.1 you get 50, if you divide 5 by 0.01 you get 500 and so on. As your denominator gets smaller and your numerator stays the same the value it generates increases until ∞, which isn't a number therefore you can't divide by 0.

**This is were my notes seem to be back on track =S.**We graphed

**abs(x) / x**and when you go

**"Trace" x=0**you see that

*y*has no value. Looking on the table there's two limits, these are called the

**right hand limit**and the

**left hand limit**. The right hand limit is at 1 and the left hand limit is at -1. Numerically it's written like this:

**lim x->0+ abs(x) / x lim x->0- abs(x) / x**If you were to have

**lim x->0 abs(x) / x**, this would work if on the graph the above limits met at the same place, however they don't so the value does not exist.

Homework for tonight, or should I say earlier was I believe EXERCISE 2.7 on The Limit of a Function. I think it was just odd number questions only. There was also suppose to be a link or something of some sort to this derivative quiz on this website but last time I checked it hadn't been up yet.

Once again I apologize if any of this seems jibberish to you, please feel welcome to comment and tell me every I missed for this class (which is probably a lot) and the next scribe will be Lindsay. It was an easy choice, she told me to pick her and me being the nice guy I am had to agree =D

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