Prelim 2016 Maths Help Thread (1 Viewer)

[eyeseeyou]

f(3)=(3-1)^2+2
=6
Sub into f^-1(x)
f^-1(6) = 1+ √((6)-2)
=1+2
=3

yes Shuuya got it right

Thanks again (+ free rep)

 

[eyeseeyou]

Why do you think I'm wrong and the answers aren't?

(These are just technicalities that the HSC doesn't really pay attention to.)

Um...b/c there are answers for a reason...

 

[InteGrand]

Um...b/c there are answers for a reason...

I think you missed the point.

I know what the Q's mean, it's just they're technically not written right. If a function is only defined for x larger than 1, then it's meaningless to sub. in a value less than 1.

What the Q's meant was something like treat f-inverse as the inverse of the function f*, where f* is the function f restricted to x >= 1, where f is the function defined for all real x by *insert formula*.

(Unless you made a typo somewhere.)

 

[eyeseeyou]

I think you missed the point.

I know what the Q's mean, it's just they're technically not written right. If a function is only defined for x larger than 1, then it's meaningless to sub. in a value less than 1.

What the Q's meant was something like treat f-inverse as the inverse of the function f*, where f* is the function f restricted to x >= 1, where f is the function defined for all real x by *insert formula*.

Uh...You make my life hard Integrand

 

[eyeseeyou]

More I need help with:

2.Given g(x)=(x+1)^2+3,x≤-1
a.g^(-1) (x)=(x+1)^2+3,x≤-1
b.g^(-1) (g(0))
c.g^(-1(g(b)) ),b>1

BTW Shuuya, if it's subbing then can you show me how you did it? (just need to check with the answers if you are doing it right or not)

Shuuya?

 

[eyeseeyou]

f(3)=(3-1)^2+2
=6
Sub into f^-1(x)
f^-1(6) = 1+ √((6)-2)
=1+2
=3

Um...could you please explain the subbing into the f^-1(x) part???

I don't get it

 

[InteGrand]

Um...could you please explain the subbing into the f^-1(x) part???

I don't get it

Did you find an expression for f^-1 (x) earlier?

 

[eyeseeyou]

Did you find an expression for f^-1 (x) earlier?

no....

 

[InteGrand]

no....

Well you'd better find it now then.

 

[leehuan]

Um...could you please explain the subbing into the f^-1(x) part???

I don't get it

You're joking...

Go back to revising functions before doing inverse functions.

Established above: f^-1(x) = 1+√(x-2)
BECAUSE we know f(3) = 6

f^-1(f(3)) = f^-1(6) = 1+√(6-2)

This is exactly what I mean by knowing your basics before you tackle the hard stuff.

 

[InteGrand]

Sent from my iPhone using Tapatalk

no....

Shuuya found it for you.

 

[eyeseeyou]

Shuuya found it for you.

I didn't realise that was a build up of the previous question...

 

[InteGrand]

I didn't realise that was a build up of the previous question...

That was probably the whole point of the previous question.

 

[leehuan]

I didn't realise that was a build up of the previous question...

............

People put questions together for a reason.

 

[eyeseeyou]

............

People put questions together for a reason.

Meh okay...I get it now

 

[leehuan]

Meh okay...I get it now

And get used to it. That's gone way beyond common in the HSC. It's something to EXPECT.

 

[eyeseeyou]

And get used to it. That's gone way beyond common in the HSC. It's something to EXPECT.

Why do you keep bolding your comments leehuan?

 

[eyeseeyou]

BTW I remember you all level SOS maths thread. I liked that thread until Trebla had to close it

 

[leehuan]

BTW I remember you all level SOS maths thread. I liked that thread until Trebla had to close it

I just made like 20 threads afterwards. Still was fine. But I think next semester I'm making a MATH12xx and discrete math thread in the uni section.

 

[Nailgun]

omg i love this thread

I can't wait until eyeseeyou starts doing 4U