BEC'S methods questions

[Mao]

and it just happens to be the point of inflection.

if you are doing spec:

the first derivative of a cubic is a quadratic, the second derivative of the cubic is then linear. when the second derivative is 0, you have a point of inflection.
the thing you should notice is that because of the symmetry of quadratic functions, the midpoint between the x-intercept is actually its axis of symmetry, or the position of its maximum/minimum. so you actually end up with the point of inflection as its midpoint.

there are better explanations, and steps showing why that is so, but you dont really need to know that.


you will see that:

1. it goes through all three of the critical points (first year concept)

2. it is the same gradient as the point of inflexion (which also happens to be the minimum/maximum gradient, depends on how the graph is)

[bec]

the thing you should notice is that because of the symmetry of quadratic functions, the midpoint between the x-intercept is actually its axis of symmetry, or the position of its maximum/minimum. so you actually end up with the point of inflection as its midpoint.

isn't this a cubic though? or because it has a squared term, does that make it share some of its properties with quadratic graphs?

[Mao]

cubic is a rotation of 180^o from its centre (point of inflection), (odd symmetry, but you dont need to know that)

the quadratic i'm refering to is the first derivative.

you will find the same thing happening to all the cubics if you do these steps

[bec]

yep yep i see what you mean:
the derivitave of a cubic is a quadratic, and the midpoint between the 2 x-ints in a quadratic is a tp, so therefore has gradient = 0....which means that the midpoint between 2 points of zero gradient (f'x=0) on a cubic is a POI - (f''x=0). is that right?
i don't do spesh btw, so i'm guessing i won't need to know any of this!

[Mao]

thats right, but you dont need to know that

the things you should notice is:
the important property is that the POI has the minimum gradient of the cubic if it is positive, and vice versa.
also, that the symmetry of the cubic is odd, that is, rotation of 180^o degrees around the POI.

[bec]

Consider the function f given by f(x)=cosx+sinx,
Find the coordinates of the points on the graph of f for which f'(x)=0

Can I do this without using the calc? Is there some identity that could help here?

[Mao]

yes you can, and you wont need identities (just a knowledge of the properties of the graphs)









which occurs at and (in the 1st and 3rd quadrant)

[Matt The Rat]

[Matt The Rat]

Bugger, beaten to the punch as usual, Mao

[/0]

Consider the function f given by f(x)=cosx+sinx,
Find the coordinates of the points on the graph of f for which f'(x)=0

Can I do this without using the calc? Is there some identity that could help here?

Not necessary, but nice to know anyways, is that:



(thanks goes to wikipedia... now I'm sure there's a way to express that without the nasty inverse tan, but I've forgotten it)

Once you've done that, the amplitude will be the y-coordinate where the gradient is 0. Then just find the x.

[bec]

thanks

[bec]

When , find f'(x) and express it in the form

I always have a lot of trouble simplifying fractions into a specific form - can anyone help me with any "general" advice on how to approach these kinds of questions?

Thanks

[Collin Li]

I'm not really sure where your problem lies. The common problem that most of my students run into is that they prefer the product rule over the quotient rule, so if it was the case , then they would use the right hand side to use the product rule, rather than the quotient rule.

In multi-choice questions (where you have to match your derivative to some particular form), I strongly recommend sticking to the quotient rule, although it really makes little difference.

I think practice will make perfect.

***

For this question, specifically, you will end up with two terms:
* one with the denominator specified in the form, and
* one without a denominator

Get a common denominator, by multiplying the term by .

[bec]

I'm not really sure where your problem lies.

My problem isn't so much in the calculus, but the fractions...you did answer my question though:

Get a common denominator, by multiplying the term by .

That makes sense, but I never thought to do that... So basically any time i'm given a form that i need to "put" something in, i just do that? I know this is year 7 stuff but somehow i've never actually been taught anything about fractions!

[Collin Li]

Okay. My point above wasn't about the calculus, it was more that if you follow the rule (i.e.: quotient, rather than product) that seems obvious, your answer will most likely resemble the form of an answer presented in the multi-choice options. This wasn't an option in this case, which is why I was a bit confused as to what was the source of your problems.