Blakhitman's Questions!

[Blakhitman]

y= - |x - 5|-6

Modulus is slightly different to other functions for transformations.

Actually, wouldn't it be ?
The reflection occurs after the translation...

YES you're right...you freaking genius!

But how did you get to that, I don't get it, so what if the reflection occurs after the translation?

I put (e) but it was wrong.

[naved_s9994]

Ahh so he is right.
Then in that case translation the reflection.

Go by whats in question, as stated by Watch.

[Blakhitman]

Oh I just tried sketching each step and yeah I get it now, but how do I do it without sketching the graph?

like I don't know how I'd end up with a positive 6 from x-axis translation if I hadn't sketched...

[the.watchman]

Oh I just tried sketching each step and yeah I get it now, but how do I do it without sketching the graph?

like I don't know how I'd end up with a positive 6 from x-axis translation if I hadn't sketched...

Whew, I'm glad I got it right

If you let ,

Then horizontal translation:

Vertical translation:

Reflection:

Hope this makes some sense!

[Blakhitman]

thanks heaps Watchman...just saw this...

ok question...how do you do this?

If has exactly one solution, find the value of correct to three decimal places.

[/0]

If then

So they are inverse functions, and are symmetrical about .
Thus, if they intersect, say at the intersection must lie on the line . Furthermore, , is concave up and is concave down, so if they only intersect once then they must 'touch' and not 'cross' - this means that their derivatives must be equal at the point of intersection. (draw this to see)

[Blakhitman]

Yeah, but how do I find the value of a? Could you do it please?

Sorry if I sound like an idiot, and if it's really simple, I'll vow to always allow a period of 2 hours to think about the question

[the.watchman]

thanks heaps Watchman...just saw this...

ok question...how do you do this?

If has exactly one solution, find the value of correct to three decimal places.

Haha, more than a month later

I tried using /0's method, but got a bit stuck, my head's still in a daze after too much music rehearsal.
Anyone else trying this one?

EDIT: Maybe try solving

AND ----->

That may work out better than what I was doing earlier, I really can't do it in this state of mind

[Blakhitman]

haha I said thanks -_-...but nah it'll help for the SAC and upcoming stuff anyway.

EDIT: I'm starting to think maybe my teacher forgot some info with this question.

[the.watchman]

haha I said thanks -_-...but nah it'll help for the SAC and upcoming stuff anyway.

EDIT: I'm starting to think maybe my teacher forgot some info with this question.

Don't think so, try subbing the first eqn into the second and solve for a, I think that's how to do it

[brightsky]

Hmm..three equations, three unknowns, should be able to get the answer...

[Blakhitman]

Can someone find the value of a then please? I still can't get it :S

[Blakhitman]

And how did you find that? just quick process no need for latex and equations and such

[the.watchman]

Hmm..three equations, three unknowns, should be able to get the answer...

Erm two, f(x)=x and f^-1(x)=x are the same

The only two you want are f(x)=x and f'(x) = [f^-1(x)]', which I have in my previous post

Can someone find the value of a then please? I still can't get it :S

I'm sorry, I would but I need to sleep

(Sorry to hijack, but if anyone is free, MYO chamber strings is doing a concert tomorrow at St John's Southgate at 3:00, would be great if any of you can make it, sorry again )

[the.watchman]

Hmmm...
I don't think you can let y=everything, as obviously the derivative graphs DO NOT follow the relation y=x, only the originals

Anyway:





(1)

Because

Therefore subbing into (1):



Subbing this into ,



Not too sure about this one though