GCSE Trial and Improvement

Discussion in 'Mathematics' started by rustybug, Feb 18, 2012.

1. rustybug

I confess I am sometimes a little uncertain about this topic. For example, the equation
x^3 - x = 20, find x to 1dp
x=2.8 gives 19.152
x=2.9 gives 21.489
From here it seems likely the answer is 2.8 (My first question: how safe is it to suggest use of linear interpolation and say that 19.152 is nearer 20 than 21.489?)
However, markschemes say at least one trial to 2dp should be shown.
x=2.85 gives 20.299
Many students would say, I was told to show one trial to 2dp, I have done it and it gives the closest answer so I shall choose 2.85 as my answer, which to 1dp is 2.9.
This leads to my second question: should you say not to try 2.85 but to use 2.84 (and 2.86 if necessary) instead and see which of those is closer to avoid the problem of 2.85 seeming to be good and leading you to give the solution as 2.9?
x=2.84 gives 20.066, which is closer than 20.299, so the solution to 1dp is 2.8.
How do you approach this?

2. rustybug

I confess I am sometimes a little uncertain about this topic. For example, the equation
x^3 - x = 20, find x to 1dp
x=2.8 gives 19.152
x=2.9 gives 21.489
From here it seems likely the answer is 2.8 (My first question: how safe is it to suggest use of linear interpolation and say that 19.152 is nearer 20 than 21.489?)
However, markschemes say at least one trial to 2dp should be shown.
x=2.85 gives 20.299
Many students would say, I was told to show one trial to 2dp, I have done it and it gives the closest answer so I shall choose 2.85 as my answer, which to 1dp is 2.9.
This leads to my second question: should you say not to try 2.85 but to use 2.84 (and 2.86 if necessary) instead and see which of those is closer to avoid the problem of 2.85 seeming to be good and leading you to give the solution as 2.9?
x=2.84 gives 20.066, which is closer than 20.299, so the solution to 1dp is 2.8.
How do you approach this?

3. AnonymousNew commenter

So 2.80 = 19.152
2.85 = 20.299
2.90 = 21.489
(which suggests the answer is between 2.80 and 2.85)
Maybe the children should be taught to think about the answer and to use their understanding of maths to think about what the answer should be rather than relying on methods.
I occasionaly use a 2dp numberline to reinforce the 1dp answer.

4. SC00BY

Your workings show that the answer lies between 2.8 and 2.9

You cannot however determine which of these is closest to the actual solution by looking at their corresponding values. You must test the mid point of the two values, hence the requirement to work out the value of 2.85, since this value is too big, ie gives a value bigger than 20, the actual value must be smaller than 2.85, therefore 2.8 is the correct one to choose!

You only ever need to go to 2dp to test the mid point of the two values.

Hope this helps.

5. DMNew commenter

As scoobs says you MUST test 2.85 (a more accurate test than this would also suffice) or you will drop one mark. It is definitely not reasonable to assume a cubic function is linear between two points.

6. DMNew commenter

If your students all have Casio Natural Display calculators it is easy to teach them how to solve these using the Table of Values feature by the way and just copy down the relevant part of the table.

7. PaulDGOccasional commenter

The reason markschemes require the 2dp trial is precisely because linear interpolation is not "safe" in the general case.

Kids find this confusing - at GCSE, the only functions they'll meet in a trial and improvement question can almost certainly be resolved with linear interpolation ("done in their heads") and so they often struggle to see the point of doing that last step as it always confirms what was "obvious" to them.

I find it's best taught as "you must always do the last step to get full marks, even though it seems really obvious, it's what the examiner requires". If you have a very able group, you can explain to them why it's important.
Then they're being taught incorrectly! - the calculation for 2.85 shows 2.85 as being above the required value, so the required answer must be 2.8 to 1d.p.

Draw number lines for them - show the curve of the function cutting the line at some point and "zoom in"... Do this with a series of powerpoints if you like, but it's even more effective if you do it with something like autograph using the slow-plot "tortoise" function and get them to predict where the datum will be cut.

8. rustybug

I have always encouraged them to buy Natural Displays, but haven't even discovered the Table of Values function! Project for the afternoon!

9. rustybug

(I think I may have mucked up my nested quotes here...)
I have just marked a set of tests and had to dock almost all of them for ignoring my instruction to give the trial at 2dp.
Which means I didn't get to test my prediction that, upon deciding 2.85 gave the best answer thus far (even including the 2dp calc that Miss says you have to do), they would round to 2.9 because they know that 2.85 to 1dp is 2.9!!
The group is bottom set - if not, I would give them some functions where linear interpolation steers you the worng way and the last (2dp) trial does not confirm what you thought obvious.
Off to figure out DM's fancy table thing now.

10. AnonymousNew commenter

When you say bottom set - what are they predicted? Do they need this or are there other things they could be focussing on that just maybe are a bit more important to a bottom set?

11. PaulDGOccasional commenter

"Mode 3"

Then it works more or less as you'd expect - you enter the equation, start & end points and step value and out comes the table.

If you want to really show off, use the A/B/C/D/E/F keys to enter coefficients and modify those coefficients by storing values in the A/B/C/D/E/F memories. Then you can get tables for curve families (one table at a time) without having to re-enter the equation.

12. NazardNew commenter

I follow all of this, but have a problem with what the original question is actually asking.

With the question that started this thread: "x^3 - x = 20, find x to 1dp", isn't is asking "what value of x makes the value of x^3 - x closest to 20?".

Here is an example: "x^3-2(x^2)+2x = 5 has a solution between 2 and 3. Find x to 1dp."

x = 2.1 gives the cubic equal to 4.641

x = 2.2 gives 5.368

x = 2.15 gives 4.993375, which suggests that, because 2.15 is too small 2.2 is the better answer.

Yet, if you look at how far away we are from the 5 that we are aiming for, x = 2.1 has an error of 0.359, while x = 2.2 has an error of 0.368. Surely this means that 2.1 is a better value for x because the value of the cubic is closer to 5 ?

Or not?

[Probability of a typo in this post? Fairly close to 1 I should think ...]

No.

14. SC00BY

2.2 is not only the better answer, it is the correct answer! If you actually solve the equation, the solution will be closer to 2.2 than 2.1 (since 2.15 is too small, the actual answer must be the bigger one).

As DM already pointed out, cubic equations do not behave in a linear way so looking how close the values are to 5 does not allow you to make any judgement as to which one is closer to the actual solution.

The closest answer correct to 1 dp is 2.2!

15. NazardNew commenter

Thanks Sc00bs - I am happier with this now. The difficulty I had was assuming it meant "what value with one decimal place ...", whereas we want the actual value for x, and then give it rounded off to 1dp, which the "find the middle value" system gives rather neatly.

Thanks to others for input too.

16. rustybug

My question (which I was too lazy to type out verbatim) said:
=====================================================
The equation
x^3-x=20
has a solution between 2 and 3. Use a trial and improvement method to find this solution. Give your solution correct to 1 decimal place. You must show ALL your working.
========================================================
if anyone is interested!
Their targets are Cs. Which is probably rather ambitious for some of them. T&I questions are always 4 marks, and I think 4 quite easy marks really as long as they have a good calc and do enough trials.
I'm actually not quite sure about what Nazard was saying, I jumped on quick to type out the question and now of course I can't see his post. Will have a closer read now. Also thanks for the advice on the tables function on the calc - unexpected guests pitched up here so I haven't had a chance to do that yet.