Monday, September 12, 2005

From the Calculus Exam

Or, "How's that plan to learn nothing in my class working for you, Mr. I-Took-Calculus-in-High-School?"

Question: Precisely state the Intermediate Value Theorem.


In a continuous function, if there are 3 points a, b, and c where a>c, there should be a point b in between a and c so that a>b>c.

Given a number a. IVT converts a to the last greatest integer a is divisible by.

Between any two points on a graph both L and the limit x -> L exist and can be formed.

If 2 numbers are substituted for x at the same distance away from c, then L is the intermediate number.

If you have a f(x) and domain (a,b) and the lim x->a f(x) = f(a) and lim x->b f(x) = f(b). If you have one point c between (a,b) then lim x->c f(x) is between f(a) and f(b).

Take the intermediate value to find where the function is discontinuous using [[x]].

The IVT states that if a function is not continuous at every number in its domain, then it is not continuous.

For every function that is continuous through the given points f(a) and f(b) and has a defined limit, there exists a real number N anywhere between the points f(a) and f(b).

On a function f(x) all values of x on the functions domain will be continuous.

lim x->a f(x) = L

In a closed interval when both f(a) and f(b) exists and are continuous, there exists a constant c between f(a) and f(b) which is also continuous.

If a function is continuous on an interval [a,b] there is at least one point c where a < c < b such that there exists f(c)=N.