Prelim 2016 Maths Help Thread (3 Viewers)

davidgoes4wce · Jun 13, 2016

$As integrand already posted the solution, you have to be able to identify that it is a Vertical Parabola with a positive coefficient for x^2$

$Think the best way to do this question is to draw in the a) Directrix first b) Draw the Focus c) Identify where that Vertex is (it is usually halfway between the Focus and Directrix$

$That vertical distance between the Focus and the Vertex is known as the 'a' constant (not sure if there is a specific name for it)$

$Knowing that this is a vertical parabola , the formula you need to remember is (x-h)^2=4a (y-k)$

$Where h represents the horizontal shift, and k represents the vertical shift. The vertex in this case is (h,k), which in this case is (2,1)$

$$ In this example, a=3, h=2, k=1 . By plugging these numbers into the equation (x-2)^2=4 \times 3 (y-1)$

$(x-2)^2=12(y-1)$

$y=\frac{(x-2)^2+12}{12}$

$As a side note, if you are ever confused about which side the parabola is facing, I always try the parabola so that it is opposite direction to the directrix. (Or another way of thinking about it is the vertex is scared of the directrix)$

eyeseeyou · Jun 13, 2016

Okay I am very confused with this:
1. Find the inverse function f^-1(x) of each function and verify that f^-1(f(x))=f(f^-1(x))=x
a. f(x)=(x-2)^2, x is less than or equal to 2
b. f(x)=x(x+2), x is greater than or equal to 1
c. f(x)=(x-1)^2+2, x is greater than or equal to 1

Green Yoda · Jun 13, 2016

davidgoes4wce said:
$As integrand already posted the solution, you have to be able to identify that it is a Vertical Parabola with a positive coefficient for x^2$

$Think the best way to do this question is to draw in the a) Directrix first b) Draw the Focus c) Identify where that Vertex is (it is usually halfway between the Focus and Directrix$

$That vertical distance between the Focus and the Vertex is known as the 'a' constant (not sure if there is a specific name for it)$

$Knowing that this is a vertical parabola , the formula you need to remember is (x-h)^2=4a (y-k)$

$Where h represents the horizontal shift, and k represents the vertical shift. The vertex in this case is (h,k), which in this case is (2,1)$

$$ In this example, a=3, h=2, k=1 . By plugging these numbers into the equation (x-2)^2=4 \times 3 (y-1)$

$(x-2)^2=12(y-1)$

$y=\frac{(x-2)^2+12}{12}$

$As a side note, if you are ever confused about which side the parabola is facing, I always try the parabola so that it is opposite direction to the directrix. (Or another way of thinking about it is the vertex is scared of the directrix)$

Great explanation

Also you can leave it in the form (x-2)^2=12(y-1) unless they specify to leave it in terms of y

davidgoes4wce · Jun 13, 2016

Also suggest for any students in Year 11 and 12 taking 2 Unit and 3 Unit, get a copy of Warwick Marlin's formulae book for Year 11 and Year 12 (2 separate books), they are about 60 pages each book. It has every formula you need to remember for the course. (The cost is $10 per book and you can get it pretty much at any Dymocks store). Even when I do the questions , I have to time from time refer to the formula book, its also a good way of refreshing your memory on key concepts.

Here is a link to the books:

https://www.understandingmaths.com/catalog

davidgoes4wce · Jun 13, 2016

I know in the HSC they now allow a formula sheet, but some of those formulas you do need to ROTE memorise as it will aid you in understanding the questions.

davidgoes4wce · Jun 13, 2016

davidgoes4wce said:
Also suggest for any students in Year 11 and 12 taking 2 Unit and 3 Unit, get a copy of Warwick Marlin's formulae book for Year 11 and Year 12 (2 separate books), they are about 60 pages each book. It has every formula you need to remember for the course. (The cost is $10 per book and you can get it pretty much at any Dymocks store). Even when I do the questions , I have to time from time refer to the formula book, its also a good way of refreshing your memory on key concepts.

Here is a link to the books:

https://www.understandingmaths.com/catalog

The books I specifically refer to for 3 Unit:

Facts and Formulas Year 11 Extension 1 (2&3 Unit)
ISBN: 9781875462216

and

Facts and Formulas Year 12 Extension 1 HSC
ISBN: 9781875462230

eyeseeyou · Jun 13, 2016

davidgoes4wce said:
The books I specifically refer to for 3 Unit:

Facts and Formulas Year 11 Extension 1 (2&3 Unit)
ISBN: 9781875462216

and

Facts and Formulas Year 12 Extension 1 HSC
ISBN: 9781875462230

How come there's no book for extension 2?

Trebla · Jun 13, 2016

eyeseeyou said:
Someone please help me with this???

Can you please answer my question?

Trebla said:
So what was the problem you had with the question originally?

Trebla · Jun 13, 2016

eyeseeyou said:
Okay I am very confused with this:
1. Find the inverse function f^-1(x) of each function and verify that f^-1(f(x))=f(f^-1(x))=x
a. f(x)=(x-2)^2, x is less than or equal to 2
b. f(x)=x(x+2), x is greater than or equal to 1
c. f(x)=(x-1)^2+2, x is greater than or equal to 1

Ok let's start with f(x)=(x-2)² for x < 2. What did you get for the inverse function f^-1(x)?

davidgoes4wce · Jun 13, 2016

eyeseeyou said:
Okay I am very confused with this:
1. Find the inverse function f^-1(x) of each function and verify that f^-1(f(x))=f(f^-1(x))=x
a. f(x)=(x-2)^2, x is less than or equal to 2

$$ I got a bit confused determining $ f^{-1}(x)=-\sqrt{x}+2$

$$ Remembering that f(x)=(x-2)^2, $ \ $ Domain $ \{x \in \mathbb{R}: x \leq 2} \} $ and Range $ \{y \in \mathbb{R} : y \geq 0 \}$

$$ I then preceded to do the inverse $ f^{-1}(x)$

$x=(y-2)^2$

$\pm \sqrt{x}=y-2$

$\pm \sqrt{x}+2=y=f^{-1}(x)$

$$ For the inverse the Domain \{x \in \mathbb{R}: x \geq 0} \} \{y \in \mathbb{R} : y \leq 2 \}$

$$ So the inverse function must be $ f^{-1}(x)=-\sqrt{x}+2$

$$ The part I had trouble with is proving $ f(f^{-1}(x))=x $ and $ f^{-1}(f(x))=x$

$f(f^{-1}(x))=[(-\sqrt{x}+2)-2]^2=(-\sqrt{x})^2=x$

$Knowing that $ f(x)=(x-2)^2 $ and $ f^{-1}(x)=-\sqrt{x}+2 $ (Are we supposed to use this one? or the +\sqrt{x}+2 \?)$

$When I did it for the inverse I came a bit unstuck, for $ f^{-1}(f(x))=-\sqrt{[ (x-2)^2]}+2= (-x+2)+2=-x+4 $ (which was my initial thinking)$

$I can see that it works for the formula when you a positive sign (which contradicts the domain mentioned in the first part for f(x)$

$+\sqrt{[ (x-2)^2]}+2 =x, $ which gives us our line of symmetry.$$

eyeseeyou · Jun 13, 2016

Trebla said:
Can you please answer my question?

Okay Gimme a min

Soz for the delay

Had stuff to do

eyeseeyou · Jun 13, 2016

Trebla said:
Can you please answer my question?

Do I answer both your Questions Trebla or what?

eyeseeyou · Jun 13, 2016

eyeseeyou said:
Okay I am very confused with this:
1. Find the inverse function f^-1(x) of each function and verify that f^-1(f(x))=f(f^-1(x))=x
a. f(x)=(x-2)^2, x is less than or equal to 2
b. f(x)=x(x+2), x is greater than or equal to 1
c. f(x)=(x-1)^2+2, x is greater than or equal to 1

BUMP

eyeseeyou · Jun 13, 2016

davidgoes4wce said:
$$ I got a bit confused determining $ f^{-1}(x)=-\sqrt{x}+2$

$$ Remembering that f(x)=(x-2)^2, $ \ $ Domain $ \{x \in \mathbb{R}: x \leq 2} \} $ and Range $ \{y \in \mathbb{R} : y \geq 0 \}$

$$ I then preceded to do the inverse $ f^{-1}(x)$

$x=(y-2)^2$

$\pm \sqrt{x}=y-2$

$\pm \sqrt{x}+2=y=f^{-1}(x)$

$$ For the inverse the Domain \{x \in \mathbb{R}: x \geq 0} \} \{y \in \mathbb{R} : y \leq 2 \}$

$$ So the inverse function must be $ f^{-1}(x)=-\sqrt{x}+2$

$$ The part I had trouble with is proving $ f(f^{-1}(x))=x $ and $ f^{-1}(f(x))=x$

$f(f^{-1}(x))=[(-\sqrt{x}+2)-2]^2=(-\sqrt{x})^2=x$

$Knowing that $ f(x)=(x-2)^2 $ and $ f^{-1}(x)=-\sqrt{x}+2 $ (Are we supposed to use this one? or the +\sqrt{x}+2 \?)$

$When I did it for the inverse I came a bit unstuck, for $ f^{-1}(f(x))=-\sqrt{[ (x-2)^2]}+2= (-x+2)+2=-x+4 $ (which was my initial thinking)$

$I can see that it works for the formula when you a positive sign (which contradicts the domain mentioned in the first part for f(x)$

$+\sqrt{[ (x-2)^2]}+2 =x, $ which gives us our line of symmetry.$$

The answers aren't on the sheet David so idk what the answers are

eyeseeyou · Jun 13, 2016

davidgoes4wce said:
$$ I got a bit confused determining $ f^{-1}(x)=-\sqrt{x}+2$

$$ Remembering that f(x)=(x-2)^2, $ \ $ Domain $ \{x \in \mathbb{R}: x \leq 2} \} $ and Range $ \{y \in \mathbb{R} : y \geq 0 \}$

$$ I then preceded to do the inverse $ f^{-1}(x)$

$x=(y-2)^2$

$\pm \sqrt{x}=y-2$

$\pm \sqrt{x}+2=y=f^{-1}(x)$

$$ For the inverse the Domain \{x \in \mathbb{R}: x \geq 0} \} \{y \in \mathbb{R} : y \leq 2 \}$

$$ So the inverse function must be $ f^{-1}(x)=-\sqrt{x}+2$

$$ The part I had trouble with is proving $ f(f^{-1}(x))=x $ and $ f^{-1}(f(x))=x$

$f(f^{-1}(x))=[(-\sqrt{x}+2)-2]^2=(-\sqrt{x})^2=x$

$Knowing that $ f(x)=(x-2)^2 $ and $ f^{-1}(x)=-\sqrt{x}+2 $ (Are we supposed to use this one? or the +\sqrt{x}+2 \?)$

$When I did it for the inverse I came a bit unstuck, for $ f^{-1}(f(x))=-\sqrt{[ (x-2)^2]}+2= (-x+2)+2=-x+4 $ (which was my initial thinking)$

$I can see that it works for the formula when you a positive sign (which contradicts the domain mentioned in the first part for f(x)$

$+\sqrt{[ (x-2)^2]}+2 =x, $ which gives us our line of symmetry.$$

The answers aren't on the sheet David so idk what the answers are

Trebla · Jun 13, 2016

Trebla said:
Ok let's start with f(x)=(x-2)² for x < 2. What did you get for the inverse function f^-1(x)?

Bump

Trebla · Jun 13, 2016

eyeseeyou said:
Do I answer both your Questions Trebla or what?

Yes. That question you managed to answer yourself, what was the problem you had with it originally that prompted you to ask it?

eyeseeyou · Jun 13, 2016

Trebla said:
Yes. That question you managed to answer yourself, what was the problem you had with it originally that prompted you to ask it?

if you are referring the question with the cubic I think that problem was that I expanded it out by accident when I wasn't supposed to

BTW Trebla, you usually don't seem to help out a lot with the maths threads

eyeseeyou · Jun 13, 2016

Trebla said:
Bump

I'll try it here

y=(x-2)^2, x <_ 2
Switch x and y around
x=(y-2)^2, y <_ 2
Squareroot of x=y-2, x <_2
squareroot of x+2=y
y=squareroot of (x) +2
Given our condition is x <_2 we can only have one case

Okay wtf am I doing wrong

Get carrotsticks to help us out

InteGrand · Jun 13, 2016

eyeseeyou said:
I'll try it here

y=(x-2)^2, x <_ 2
Switch x and y around
x=(y-2)^2, y <_ 2
Squareroot of x=y-2, x <_2
squareroot of x+2=y
y=squareroot of (x) +2
Given our condition is x <_2 we can only have one case

Okay wtf am I doing wrong

Get carrotsticks to help us out

You needed to take the negative square root, since y needed to be ≤ 2.

Prelim 2016 Maths Help Thread (3 Viewers)

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