I am emailing a link of this to everyone on the class list every Friday afternoon. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.ie and I will add you to the mailing list.


These are finally corrected. I would like to apologise for the inexplicable delay… you are identified by the last five digits of your student number. Remarks below.

S/N Mark Ex 12.5
27898 12.5
29579 12.5
89138 12.5
35809 12
00988 12
05441 12
72936 11.5
31148 11
59663 11
89362 11
19801 10.5
76939 10
55503 10
64923 10
50316 10
56576 9.5
01642 9.5
81431 9.5
28745 9.5
25441 9
11938 9
93481 9
40198 9
37211 9
28575 8.5
30609 8
09341 8
98786 8
04996 7.5
21361 7.5
21967 7
29768 7
00633 7
59593 6.5
84181 6
35726 6
32338 5.5
07743 5.5
30948 5.5
74522 4.5
47692 4.5
93528 4
28475 4
59528 3.5
15585 3.5
71579 3
09658 3
47796 5
64301 5


Firstly everyone who handed up clearly put in a massive effort. I think twelve students didn’t hand anything at all in.

Project 5 (a)

This was the project where you had to justify the steps taken in the solutions of the relevant questions. Omissions (not justifying steps), errors and superfluities/waffle were penalised here.

Project 5 (b)

This project was marked on

  • Length (3 marks) — all but three students scored full marks. These students lost marks for not doing all the questions or very ‘light’ worked solutions.
  • Presentation (2 marks) — four students got one mark; everyone else got full marks. Worked solutions aren’t going to much use if you can’t read them easily.
  • Accuracy (4 marks) — basically I counted the number of errors in the worked solutions and worked from there. We had from 4 to 38 errors in our worked solutions. I gave 4/4 for <8 errors, 3/4 for 9-12, 2/4 for 13-16, 1/4 for <24 errors and 0/4 for more.
  • Algebra (3 marks) — marks for algebra explanations and justification of steps. I came down heavily on algebra errors or stuff like “the square and the square root cancel”. Also writing stuff like \sin \Rightarrow \cos and similar nonsenses.
  • Explanations (5 marks) — there were a few things that needed explaining. Note that giving the steps in the method is not explaining things at all — maths is NOT a series of methods and steps to learn off and I roundly reject these ‘explanations’. Only two students had proper explanations of all of these while twelve students gave little or no explanations of what we were doing and they only got 1 or 0/5.  The things that needed to be explained, the ideas were
  1. Implicit Differentiation
  2. Differentiation from First Principles
  3. Chain Rule
  4. Asymptotes – both vertical and horizontal
  5. Maxima and Minima
  6. ‘Roots’
  7. Increasing Function
  8. Completing the Square
  • Calculus (5 marks) — your handling of these concepts as well as the product, quotient and chain rules. How well you demonstrated that differentiation is about slopes of tangents, etc. These marks were broadly correlated with the Explanation marks.
  • Pictures (3 marks) — ever hear the phrase, a picture paints a thousand words? Some of us had only one picture while others had upwards of 11. The whole theme of calculus is looking at geometric objects in an algebraic way.
  • Language & Level (3 + 3 marks) — most people got 3/3 for language. If language was not precise or contained lots of clumsy expression I docked marks. Level means not talking about epsilon-delta to a class of Leaving Cert students for example… things should be made as simple as possible — but no simpler. Over- and under-complicated worked solutions were punished although again, most students got 3/3.
  • Overall (6 marks) — my own impression of what your work was worth. 1/6 — very poor [6 students], 2/6 poor [5 students], 3/6 fair [2 students], 4/6 good [5 students], 5/6 very good[6 students], 6/6 outstanding [3 students].

Projects 4 & 8

Only two students did these projects so I will not make comments on them here.

Project 6: L’Hopital’s Rule

Most students answered the first question well by appealing to a diagram of the graph of y=f(x) near a point x=1. The second part, which was worth 20 marks out of 75, was typically not answered very well. Either students had erroneous arguments or failed to use the linear approximation which was necessary. After this students primarily lost marks for not checking that the conditions for L’Hoptital’s Rule were met before applying it.

Project 9: Least Squares

Only two students did this projects so I will not make comments on them here.

Review Lecture

We will have a review on Monday 29 April at 10:00 in WGB G14. In this lecture I will take any questions and if we run out of questions I will detail the format of the exam. It will be same as 2011 and 2012. I am going to make Q.3 a little longer and Q.4 a little shorter.


When ye come to it, I recommend for revision that ye look at the 2011/12 Summer & Autumn papers. Do not learn off solutions! This will almost certainly not work well. Instead try and understand the concepts behind what we are doing. If you have any questions please ask me here on the webpage.


I will hopefully get a chance to look at the feedback forms and hopefully form a response.

Autumn 2012 Q. 2

Apologies for the delay but I never forgot! Is it just the first bound that you were asking about?

Using the Closed Interval Method, or otherwise, fi nd positive constants M and N
such that the following inequalities hold for all values of x in the specifi ed range:

  • |5-|2x+1||\leq M for -1\leq x\leq 1
  • |x^2-2x+7|\leq N for -4\leq x\leq -2

Use the upper bound from part (ii) and the \varepsilon\delta defi nition of a limit to prove that:

\lim_{x\rightarrow -3}(x^3+x^2+x+1)=-20.

Solution: For the first bound, it is actually not too hard to use the triangle inequality:

|5-|2x+1||\leq |5|+|-|2x+1||=5+|2x+1|

\leq 5+|2x|+1=5+2|x|+1=6+2|x|\leq 6+2(1)=8.

Can we use the Closed Interval Method though; alternatively? Let f(x)=5-|2x+1|. The Closed Interval Method says that the absolute maximum/minimum of a continuous function defined on a closed interval [a,b] is attained at a critical point:

  • the end points x=a,\,b
  • points where f'=0
  • points where f' is undefined

In the examples we have done before there was only critical points of the first two kinds. This, however, has a non-smooth point where the absolute value is zero… where it looks like a V and f' is undefined. That is f' is undefined when 2x+1=0\Rightarrow x=-1/2. The harder part of this is to find the derivative away from x=-1/2 — and to make sure that it is not zero. One option is to plot the function. Algebraically I would probably say something like this… when 2x+1<0, which is when x<-1/2, we have |2x+1|=-(2x+1). So when x<-1/2 we have f(x)=5-|2x+1|=5-(-(2x+1))=6+2x. Conversely, when 2x+1\geq 0, which is when x\geq -1/2, we have |2x+1|=2x+1 and we end up with f(x)=4-2x when x\geq -1/2. In in all then we have

f(x)=\left\{\begin{array}{cc}6+2x & \text{ if }x<-1/2\\ 4-2x & \text{ if }x\geq -1/2\end{array}\right.,

\Rightarrow f'(x)=\left\{\begin{array}{cc}2 & \text{ if }x<-1/2\\ -2 & \text{ if }x> -1/2\end{array}\right.;

in other words f'\neq 0. Hence we evaluate

  • f(-1)=5-|2(-1)+1|=5-1=4
  • f(1)=5-|2(1)+1|=2
  • f(-1/2)=5-|2(-1/2)+1|=5

Hence \max |f(x)|=5 so we can take |5-|2x+1||\leq 5=M.

To be honest if you just looked at -1,1 and -1/2 I was giving full marks. Note that the Closed Interval Method gives the best bound possible.

The second bound is routine as is the limit.

Very important: although it is actually part (ii) that you need to do the limit, there is an issue with failing to find a bound in the first part. Most of these questions look like

  • find an upper bound M
  • use this upper bound to find a limit

If you fail to find an explicit bound in the first part, you can still do the second part though. Just let your bound equal ‘M’.