The solutions to the practice problems for Exam 1 have been posted and can be obtained by clicking here.

Wednesday in Calc I, we introduced two problems that seemed to have nothing in common, but whose solutions used exactly the same method.

The tangent problem asked the question of how we can compute the slope of a curve at a particular point, since the slope changes from point to point (unlike with a line). We can reduce the problem to finding the slope of a particular line -- a tangent line. This is a line that (at least locally) only touches a curve at our point of interest. BUT ... we only know one point on the tangent line, and we generally need two points to determine a line's slope.

Another problem is one of velocity. Can we determine the velocity of an object at a particular instant? Normally we think of velocity in terms of the formula: (change in displacement) divided by (change in time). The problem is that time doesn't change during an instant.

It turns out we can use limits in both cases to sneak up on an answer. (Limits are a great way to sidle up to an answer when our usual "full-frontal assault" technique won't work.) Given a function f and a value of interest, a, we can figure out both the slope of the tangent line at x = a and the instantaneous velocity by way of the expression



By itself, the quotient represents both , the slope of the secant line between the x-values a and a + h and the average velocity over the time interval . It's by making the difference in the x-values -- or the length of the time interval we're measuring -- go to 0 that we achieve our goal.

We looked at a couple of examples for the tangent problem (). I then commended to you the job of reading up on examples of the velocity problem.

I was supposed to have talked about the Intermediate Value Theorem ... remembered in the 10:50 class, but forgot in the 7:50 class. I'll write about it when I've updated everyone.

See you at the test on Friday.

Monday we talked about the idea of continuity. A function f is continuous at a point c if



This entails more than we might think. For this equation to hold, we need (1) the limit on the left-hand side of the equation to exist; (2) the function f to be defined at c and; (3) both results to equal each other.

A function that is not continuous at c is discontinuous at c. There are a few types of discontinuities, which we classify based on how seriously the function is "broken" at the point of interest. The function has a vertical asymptote at x = 0; the discontinuity there is essential (that is, unfixable). Another unfixable discontinuity is a jump discontinuity. The greatest integer function has a jump at every integer and has discontinuities there as well.

There are, however, discontinuities that can be fixed. These are called removable singularities. Graphically, these show up as open holes, places where the function would take on the indeterminate value 0/0. For example, the function



is discontinuous at x = 2 since f is undefined there. However, the limit as exists:



So we can "fix" our function f (i.e., remove the singularity) by re-defining:



On Wednesday, we'll talk about the Intermediate Value Theorem -- one of three "value theorems" we'll encounter -- and move on toward derivatives. See you then.

Practice problems for Exam #1 can be obtained by clicking here. The rule is that you get up to 5 points on Exam #1 for a good effort. Less-than-good efforts will get less than 5 points. Feel free to ask me questions or to work with friends on these.

I will post the solutions here on Thursday, September 18th.

The first full-class exam will take place on Friday, September 19th. It will cover all we know about limits as well as some pre-calculus stuff -- Chapters 1 and 2. Stay tuned for more details.

On Friday, we went over sections 2.5 and 2.6, which discussed limits involving the transcendental functions: trigonometric functions and exponential and logarithmic functions.

The good news was that all of these functions, as long as we stay away from any "bad" x-values, work with our "plug-in" method for limits. We learned about two special limits involving the sine and cosine functions as we approach x = 0. We also looked at the limits at plus and minus infinity for exponential functions, as well as the limit at infinity for logarithmic functions (assuming the base is greater than 1). Logarithmic functions have a vertical asymptote at x = 0, so they have a limit approaching 0 from the right.

We took a look at the most commonly used base for exponential and logarithmic functions, the natural base e.



We found that

,

a fact which we sort of learned back in the days when we looked at compound interest problems.

We'll talk about continuity on Monday.