Precal Blog

During class on Friday, we spend the hour solving the problems in exercise 2.1 of the exponential functions packet. We will have some class time on monday to continue and claims will be on Tuesday.

Some tips for solving exponential function problems:

-know how the equation y=ab^x works and understand the restrictions: a cannot equal 0, b cannot equal 1, and b must be greater than 0

-understand growth rate (next/current) and relative rate (the change in percentage of the amount present)

-know that Euler’s number is the limiting value of exponential functions and know how to use it in problem solving

-create charts like the ones in the packet if you get stuck

-also, table 2.4 may be helpful in understanding the nature of change in certain situations in exercise 2.1

Good luck! The next scribe will be Emily M.

Today in precalc, we spent most of the time working on problems out of a packet that deals with exponential functions. Because we learned the basics about exponential functions yesterday, that means that today we applied these concepts.

The problem we had to solve today was to determine if there was a limit to the amount of money that would be earned in a savings account if we continually added a compound interest of 100%. The expression given to us (a form of an exponential function) was (1+1/n)^n.

First we had to make a table.

If you look at the last few values, you might notice that they have seemed to reach a peak at about 2.72. This value is actually an irrational number (like pi) that is called Euler’s number. The symbol for this value is e. The exact value (or as close as I can get) is 2.718281828… The formal definition of e is (1+1/n)^n.

So when applying this concept to banking, this means that if you want to make the most amount of money by having the interest compiled more frequently, there is actually a limit at which the final amount will not increase at a significant rate.

Basically, today was a day where we worked on a few problems and continued our study into exponential functions. The next scribe will be…Shannon. Be happy. You have the entire weekend.

Today in Pre-Calc, we spent most of the class working on Activity 2.2 in our packets, problems 1-3. These problems dealt with another type of exponential function:

f(x)= (1+1/x)^x

The first problem dealt with a real life application problem dealing with compound interest. We were told to complete a table for the function:

f(x)=100(1+.08/x)^x

We needed to figure out the values for your balance annually, semiannually, quarterly, monthly, daily, and hourly. Here are the values for each of the time periods:

Annual: $108.00; Semiannual: $108.16; Quarterly: $108.24; Monthly: $108.30; Daily: $108.33; Hourly: $108.33

The basic pattern in this function would be it increases at an exponential rate but after a certain time period, it will begin to level off. The number that this function begins to level off at would be $108.33. These types of functions are used quite frequently by almost every banker.

To figure out how to get there, we built up the function. We started off by creating a table and a graph for the function:

f(x)=1/x

That graph made an L-shape in the first quadrant. There was a horizontal asymptote at the y=.01 line. This is because the f(x) value will never get to .01, it will just approach it. Then, we derived another function, which was:

f(x)=1+1\x

This graph also produced an L-shaped curve in the first quadrant. However, the horizontal asymptote instead of being at .1, was at 1.01. This was because it will approach 1.01, but never quite reach that value. This leads us to the function I mentioned earlier:

f(x)=(1+1/x)^x

Unlike the first two graphs, this one produced a half-upside down bowl shaped curve. Sorry for the explanation of the shape. Anyways, this graph is shaped like this because your number will start low and gradually climb higher due to the exponent. The “limit” of this function is approximately 2.718….. This is actually a special number many bankers use, and its symbol is e.

On a side note, we also spent a good part of the class period talking about the new iPad and doing some random background checks. Fun times. Anyways, tomorrow’s scribe is ……………….. Emily M.

Today in honors Pre-Calculus hour 3, we discussed exponential functions. First we stated the basic equation for exponential functions can be written as f(x)= a*b^x. We further determined that exponential functions have certain “rules” they must follow, they are b>0, b can’t equal 1, and a can’t equal 0. For the first rule, if we use a number less than 0, like -2, our answers for the f(x) value would alternate from positive to negative. For the second rule, if b equals 1, we would simply get back the a value leaving us with a straight line. For the final rule, if a is equal to 0, our answer is 0 giving us nothing. With these rules in mind we looked at four different exponential graphs. I don’t have images so I’m sorry you must bear with my descriptions. The first graph was a>0, b>1. The graph was a positive exponential graph starting above the x-axis getting larger. The second graph was a<0, b>1. The graph was negative starting below the x-axis and sloping down into the fourth quadrant. The third graph was a>0, 0

Next we went over the 2.1 packet problem 3. We decided the domain for an exponential graph could be all real numbers, but the range had some stipulations. If a is negative, y<0; if a is positive, y>0. After testing the f(-x)=f(x) and f(-x)=-f(x) equations for even and odd symmetry, we decided that exponential graphs fit neither so they have no symmetry. Next we deduced that exponential graphs can increase or decrease everywhere, across the entire domain. Finally we looked at the concavity of exponential graphs. graphs can either can be concave up (“u-shaped” so they hold water) or concave down (upside down “u” so they shed water). In the case of exponential graphs if a>0, the graph is concave up; if a<0, the graph is concave down.

The last concept we discussed was “end behavior”. I’m not entirely clear on this concept (Mr. B said he struggled with it in high school so I don’t feel that bad) but basically it evaluates towards what “number” the graph is going. In the case of graph one from above the x-values are “going towards” positive infinity; and the f(x) values are also “going towards” positive infinity. When analyzing end behavior though you must make two sets of…”it”. One for the positive side and one going towards the starting point. So we have our positive side, now we look at the side going towards our starting point. in this case of graph one we have our x-values going towards negative infinity, and f(x) values going towards zero. Sorry if that was confusing, math terminology is beyond me so that’s my best attempt. We finished class by getting page 89 of the packet and to work on those problems. The next scribe will be Jake.

Today in class we talked more about Exponential functions, and went through problem 3 from the worksheet we got yesterday.
You can graph Exponential functions in four different ways:

F(x) = ab^2  <—— general equation

The b-value represent the stretch
b > 0
b -> not equal to 1
a -> not equal to 0

Domain/Range
D: all real numbers
R: if a > 0 then y > o
     if a < 0 then y < 0

Symmerty –> None
* no even symetry
* no odd symerty

Increasing/Decreasing
* always increasing or always decreasing
* always one-to-one function

Concavity
If a is positive = concavity up
If a is negative = concavity down

Inflection points:
Its the points where it change from concavity up to concavity down. And Concavity down to concavity up.

end Behavior
( in the picture on top)

The next scribe is Riley

Today we started second semester with new seats and new “textbooks.” We worked in groups on section 2.1.  It  focused on exponential functions and how different types of values compare.

I am going to show the first few values on the table for it took make more sense.

First differences (The differences between successive pairs of values) in exponential functions increase, but not at a constant rate.  If you compare linear, it is consistent, but exponential isn’t (see below picture).

The values verses the first differences is (value/first difference) and is constant.  This means the ratio is constant and it is PROPORTIONAL.

The Ratio between successive values (next value/current value) is a constant.  It has a constant growth rate.  All exponential functions had an add-multiply property. This is seen in the table below:

The 1.25 was explained.  We, in this case, wanted to increase by 25%.  We would take the value, we’ll say 200, and multiply it by .25 and add that to 200.  That leaves us at 200 + 200(.25).  This could be simplified to 200(1 +.25).  That’s how we get value times 1.25.

The next scribe is GEORGE!!!!!!!

Today we watched the film Hard Problems. The director of the film George Csicsery hopes that it will “inspire others to see mathematics as those in the film already do—the most challenging and the most rewarding of pursuits.”

I’d like you to react to what you saw. The following are some quotes from the film and some questions that may help you focus your reaction.

• “It takes a lot of stupid ideas to generate a good one”
• “Whenever I get something by my own effort is when things get really exciting”
• “[Math contests were] the first time I really discovered the subject of math rather than being taught random stuff”
• “Math is all about exploring. Trying new methods and techniques.”
• “I love not only getting the right answer but searching for the right way to get the right answer.”

What was your impression of what made the students in the film successful in mathematics?

What was your impression of how the students in the film viewed mathematics?

Here is a link to the problems that the students completed in the 2006 IMO.

Well, this is officially my second time writing this same post out. This would’ve been up sooner, but my computer crashed on me right when I had finally finished, so I had to rewrite the whole thing. But that’s besides the point.

After a slight delay, we began class with going over the properties for sine and cosine we went over yesterday.

1.) sin(A-B)= cosBsinA-cosAsinB
2.) since(A+B)= cosBsinA+cosAsinB
3.) sin2A= 2cosAsinA
4.) cos(A-B)= cosBcosA+sinBsinA
5.) cos(A+B)= cosBcoA-sinBsinA
6.) cos2A= cos^2A-sin^2A