…but certainly for Test 2!
On Monday we wrote down the definition for a function to be continuous on a closed interval $I\subset \mathbb{R}$. This is to account for limits such as $\lim_{x\rightarrow 0}\sqrt{x}$
i.e. in this case the left limit does not exist as the function is not defined for $x<0$. We wrote down the Intermediate Value Theorem.
On Wednesday we looked at the Intermediate value theorem again. For a proof see http://en.wikipedia.org/wiki/Intermediate_value_theorem We discussed how it implies that, on a closed interval, a continuous function is bounded. If the function is unbounded then it is not continuous. We showed how the Intermediate Value Theorem can estimate the location of roots of functions. Finally we showed with an example the restriction (in the theorem) to closed intervals is necessary.
We then motivated differentiation: why do we do it? We derived the formula for the derivative of a function, defined a differentiable function (at a point) and showed that, as expected, the derivative of a line $y=mx+c$ is just $m$. This also implies that a constant function has derivative zero. We did one more example (a quadratic). Finally we said that in some sense a differentiable function must be somewhat ‘nicer than’ or ‘as nice’ as a continuous function – as all differentiable functions are continuous. We will prove this on 1/11/10.

Problems

You need to do exercises – all of the following you should be able to attempt. Do as many as you can/ want in the following order of most beneficial:

Wills’ Exercise Sheets

Past Exam Papers

Q. 4(b), 4(a) from http://booleweb.ucc.ie/ExamPapers/exams2008/Maths_Stds/MS2001Sum08.pdf

Q. 3(a), 5(a) from http://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/ms2001s2004.pdf

Q. 4(b), 6(a) from http://booleweb.ucc.ie/ExamPapers/exams/Mathematical_Studies/MS2001.pdf