Will we be forced to use sgn(x) or will hybrid functions do?

[3Xamz]

http://www.mav.vic.edu.au/files/general/Solutions-to-VCAA-2010-Sample-Questions.pdf

Question 12)b)
They've given it in both ways, but hybrid should be adequate right?

I don't get the sgn(x) way of putting the answer.. Would they ever ask us to show it in sgn(x) explicitly?

[schnappy]

I've only ever seen hybrid functions in VCE. You could easily check the study design though.

[m@tty]

The signum function is NOT part of the methods course.

But in a hope of enlightening you, it simply takes the 'sign' of the expression it acts on:

Eg.





Also,


So:
.

By definition, , the signum provides the sign, while the modulus provides the magnitude.


You should leave this alone for Methods, rather splitting into cases where appropriate. =)

[3Xamz]

OMG! So that's what it means.
Im like wtf why is either -1,0 or 1 regardless of the function?

I thought the derivative was either one of those three values at any point which was obviously incorrect.

BUT THAT MAKES SO MUCH SENSE!
Thanks M@tty

[m@tty]

BUT THAT MAKES SO MUCH SENSE!

As all maths does, in the end.

No problem

[mugiwara]

sorry noob question, but when you diff that hybrid function, shouldn't the derivative not include the x = 0

[m@tty]

Indeed, you are correct, cusps and endpoints are not differentiable - they should have used a "less than" sign, rather than less than or equal to.
Good pickup; certainly not a noob question

EDIT: Sorry for the misleading response =S See later post.

[mugiwara]

but on the sample paper itself, it says "The hybrid function f is differentiable at x = 0 with f ′(0) = 0, so f ′ can either be specified as a hybrid function, or alternatively
using the sgn function, as indicated in the answer, in which case
f ′(x) = 2x sgn(x)." Are they saying that x=0 is included?

[m@tty]

Very sorry for my prior response, it is indeed differentiable at zero, as the curve is smooth and continuous.

Approaching x=0 from the left, y=0 and y'=0.

Approaching x=0 from the right, y=0 and y'=0.

Because both are the same, this is why it is smooth and continuous and thus is included in the derivative function.

Their answer stands.

[mugiwara]

oh i see now. Thanks a lot Matty. So a function is smooth if it is continuous and has no cusps?

[m@tty]

If there is a cusp - where the gradients are not equal approaching a point from opposite sides - then it is not differentiable and must be excluded from the derivative function.

If it is not continuous, ie. the y-values are different approaching a point from different sides, then it is not differentiable, just like end points.

[mugiwara]

ohhhh it's all clear now. Thanks again Matty

[m@tty]

No problem. Good luck for tomorrow =]

[doter]

this is probably a very stupid question
but do all hybrid functions differentiate into an answer with a modulus function
so confused

[3Xamz]

No, I believe the hybrid functions come from the modulus function.

The modulus function spits out two equations for two domain intervals.
So, when you derive two equation you get two gradient equations.

Once you get the two correct functions initially, you can forget about the modulus function. I do anyways, because you no longer need it as you have two functions in more familar terms to yourself, which represents the modulus function.