VCE Methods Question Thread!

[yawho]

how can one tell the approximation is correct to 5 dp?

[ashoni]

Simplify
(x - 1)^(1/2) - x/(x - 1)^(1/2)

could you show the working out please?

[ligands]

how can one tell the approximation is correct to 5 dp?

if you're referring to the question i asked, the question which i was referring to asked for 5 dp's

[yawho]

how can one tell the approximation is correct to 5 dp?

if you're referring to the question i asked, the question which i was referring to asked for 5 dp's

your question said the approximation 'correct' to 5 dp. It did not ask for 5 dp.

[ligands]

how can one tell the approximation is correct to 5 dp?

if you're referring to the question i asked, the question which i was referring to asked for 5 dp's

your question said the approximation 'correct' to 5 dp. It did not ask for 5 dp.

i wrote down the question as it was asked in the practice sac i was doing.. lol

[tony3272]

Simplify
(x - 1)^(1/2) - x/(x - 1)^(1/2)

could you show the working out please?

[TrueTears]

how can one tell the approximation is correct to 5 dp?

I don't see how you can't? It's doable by hand albeit very tedious.

Limme finish off the question then,

Then just convert to a decimal with 5 dps using long division which you should have learnt back in year 7. I would show it in this post, but I can not type out long division with latex.

It is tedious and pretty boring, but if the question asks for it, it is certainly doable by hand using the above steps.

[kamil9876]

Perhaps what yawho was asking for was: how can one tell that the approximation is correct to 5 decimal places? By doing long division you find the approximation correct to 5 decimal places, but it's not obvious that this actual approximation is correct up to 5 decimal places of the true value.

[TrueTears]

Well you can't, but for the purpose of this VCE question I'm pretty sure that's the correct interpretation, why make a fuss over pointless technicalities in the wording?

[kamil9876]

I don't know what your point is. Maybe you didn't read my post properly, this is not a technicality in the wording but a pretty fair question to ask (i.e how can you be sure that your approximation is close enough)

[TrueTears]

My point is if you want the actual approximation correct to 5dp, then get a calculator out and do it, otherwise I don't see the problem, clearly this question is a non calculator question that often appears on tech-free SACs and you're in VCE to maximise marks; not to fuss over the wording during a SAC. Yes it may not be a triviality from where you're coming from, but in the context of this question, what you are pursuing is indeed quite trivial.

EDIT: yawho you have been perma-banned many times before, and have been trolling subtly for quite a while, including your recent posts in the chem boards: Re: sig figs as a warning, please stop being a smartarse without the intention of actually helping.

[Jazz_Blue]

Can someone explain to me how to sketch this kind of graphs?



1)  f:R=R
f(0)= 0,    f'(0)=0
f(4)= 0,    f'(3) = 0
                f'(x) < 0 for ( - infinity, 0 ) U ( 0, 3)
                f'(x) >0 for ( 3, infinity)

2) f(0)= 0,    f'(0)=0
f(4)= 0,    f'(3) = 0
                f'(x) < 0 for ( x : x > 3)
                f'(x) >0 for ( x : x < 3) \ (0)

[tony3272]

Can someone explain to me how to sketch this kind of graphs?



1)  f:R=R
f(0)= 0,    f'(0)=0
f(4)= 0,    f'(3) = 0
                f'(x) < 0 for ( - infinity, 0 ) U ( 0, 3)
                f'(x) >0 for ( 3, infinity)

2) f(0)= 0,    f'(0)=0
f(4)= 0,    f'(3) = 0
                f'(x) < 0 for ( x : x > 3)
                f'(x) >0 for ( x : x < 3) \ (0)

For the first one:

You have f(0)= 0 and f(4)=0, which implies your intercepts are at x=0 and x=4.

you're also given stationary points at x=0 and x=3.

So in order to sketch it, you need to look at where the graph has a positive gradient or a negative gradient.

For f'(x) < 0 for ( - infinity, 0 ) U ( 0, 3) you can see that you're graph will have a negative gradient from - infinity to zero, then it will intercept (0,0) where we now there is also a stationary point, and then continue with a negative gradient.
In order for this to happen, there must be a stationary point of inflection at x=0.

Your graph then continues with a negative gradient until x=3, where there is a stationary point, and then becomes positive until infinity.
Therefore, you will have a local minimum at x=3, and the graph will then continue with a positive gradient through (4,0).

[Jazz_Blue]

1)  f:R=R
f(0)= 0,    f'(0)=0
f(4)= 0,    f'(3) = 0
                f'(x) < 0 for ( - infinity, 0 ) U ( 0, 3)
                f'(x) >0 for ( 3, infinity)

2) f(0)= 0,    f'(0)=0
f(4)= 0,    f'(3) = 0
                f'(x) < 0 for ( x : x > 3)
                f'(x) >0 for ( x : x < 3) \ (0)
[/quote]
For the first one:

You have f(0)= 0 and f(4)=0, which implies your intercepts are at x=0 and x=4.

you're also given stationary points at x=0 and x=3.

So in order to sketch it, you need to look at where the graph has a positive gradient or a negative gradient.

For f'(x) < 0 for ( - infinity, 0 ) U ( 0, 3) you can see that you're graph will have a negative gradient from - infinity to zero, then it will intercept (0,0) where we now there is also a stationary point, and then continue with a negative gradient.
In order for this to happen, there must be a stationary point of inflection at x=0.

Your graph then continues with a negative gradient until x=3, where there is a stationary point, and then becomes positive until infinity.
Therefore, you will have a local minimum at x=3, and the graph will then continue with a positive gradient through (4,0).

[/quote]


How to know the graph has negative or positive gradient by looking at the question? is it because of the < / > sign?

[Phy124]

How to know the graph has negative or positive gradient by looking at the question? is it because of the < / > sign?

If f'(x)>0 then the gradient is positive
If f'(x)<0 then the gradient is negative