Prelim 2016 Maths Help Thread (1 Viewer)

[leehuan]

I got y=1+ Squareroot of (x-2)

Is that the answer?

I said already the inverse function looks right

 

[Shuuya]

So I did this and am so confused:

f(x)=(x-1)^2+2,x≥1
y=(x-1)^2+2,x≥1
y-2=(x-1)^2,x≥1
x-2=(y-1)^2,y≥1
(y-1)^2=x-2,y≥1
y-1=√(x-2) ,y≥1
y=√(x-2)+1 ,y≥1

Could some1 tell me what I need to do next?

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[eyeseeyou]

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Thanks Shuuya

 

[eyeseeyou]

Okay I am dead struggling to understand wtf is going on in these questions:

1.given f(x)=(x-1)^2+2,x≥1
a.f^(-1) (f(3))
b.f^(-1) (f(-1))
c.f^(-1) (f(a)),a<1

 

[Shuuya]

Okay I am dead struggling to understand wtf is going on in these questions:

1.given f(x)=(x-1)^2+2,x≥1
a.f^(-1) (f(3))
b.f^(-1) (f(-1))
c.f^(-1) (f(a)),a<1

Just look at what I did at the end and sub in the numbers for the x's

 

[leehuan]

Okay I am dead struggling to understand wtf is going on in these questions:

1.given f(x)=(x-1)^2+2,x≥1
a.f^(-1) (f(3))
b.f^(-1) (f(-1))
c.f^(-1) (f(a)),a<1

Don't rush to do both, evaluate them one at a time

f(3)=(3-1)^2+2

Then chuck whatever value you get into the f^(-1)(x) formula


Note: There is actually an easier way out using the f(f^-1(x))=x trick but right now just do as I say.

 

[eyeseeyou]

brb

 

[eyeseeyou]

Just look at what I did at the end and sub in the numbers for the x's

no these are a different type of question(s) (I think) and you are required to draw a graph

 

[Shuuya]

no these are a different type of question(s) (I think) and you are required to draw a graph

It just looks like subbing in to me...

 

[InteGrand]

Okay I am dead struggling to understand wtf is going on in these questions:

1.given f(x)=(x-1)^2+2,x≥1
a.f^(-1) (f(3))
b.f^(-1) (f(-1))
c.f^(-1) (f(a)),a<1

 

[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)

 

[eyeseeyou]

The answers were 3,3, 2-a

 

[InteGrand]

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)

 

[eyeseeyou]

answers were -2,-2 and -2-b

 

[eyeseeyou]

BTW Integrand you'll need to draw a graph for these Q's

 

[InteGrand]

What the answers say have no impact on the statements I made.

 

[eyeseeyou]

What the answers say have no impact on the statements I made.

Wel...you're wrong (I think)

 

[Shuuya]

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)

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

 

[InteGrand]

Also I think you typoed Q. 2 a), because it's the same as the line above.

 

[InteGrand]

Wel...you're wrong (I think)

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.)