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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: AirLandBus on July 11, 2014, 12:28:20 pm

Title: Methods 1/2 Question.
Post by: AirLandBus on July 11, 2014, 12:28:20 pm: Hey guys, just need some help finding the inverse of a function for a holiday assignment. The function is x^2+4x-5 and the question asks to find the inverse of the graph (the equation). The inverse that i found im not sure is right. Thanks.
Title: Re: Methods 1/2 Question.
Post by: dankfrank420 on July 11, 2014, 01:40:01 pm: Are you sure there is no restriction on the domain or range?
Title: Re: Methods 1/2 Question.
Post by: AirLandBus on July 11, 2014, 01:46:56 pm: Nope.
Questions.
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So for a i got x^2 + 4x - 5
b i drew it.
c i drew the inverse, which i got as y= sqroot of (x+9) +2 by switching x and y and solving for y (someone check this please?)
d i have no clue?
Title: Re: Methods 1/2 Question.
Post by: dankfrank420 on July 11, 2014, 02:59:16 pm: I got the same two equations for the inverse as you did. The graphs you drew on your CAS should be somewhat perpendicular to eachother, with the inverse heading towards infinity of x and the original function heading towards the infinity of y.

Edit: I didn't really explain it that well.

We know a graph is a function if it passes the vertical line test. Obviously, our inverse does not pass it. To make the inverse a function, we must restrict its range, and to do this we must restrict the domain of the original (as domain of original = range of inverse). So pick two points and restrict the domain of the original. This will restrict the range of the inverse, therefore making it a 1-to-1 graph (and therefore a function).
Title: Re: Methods 1/2 Question.
Post by: keltingmeith on July 11, 2014, 06:59:53 pm: Quote from: AirLandBus on July 11, 2014, 01:46:56 pm
c i drew the inverse, which i got as y= sqroot of (x+9) +2 by switching x and y and solving for y (someone check this please?)

I'm just doing to nit-pick here - they didn't ask you to find the rule, so you don't have to. What you've done is right (except it should be -2, not +2), but you can do it with less work. An inverse is reflected over the line y=x - to save time and be quick about it, you should dot in the line y=x, and then draw the inverse so that it looks like it's reflected over that line. To see an example of this, draw some graphs and their inverses (knowing the equation of both), and dot in the line y=x to see what they look like. You should draw the y and x axis so that they have the same scale (which means if you do them on your CAS, you'll need to resize it)
Title: Re: Methods 1/2 Question.
Post by: idontknow2298 on July 13, 2014, 12:13:42 am: Please help me with this question, I'm struggling a lot with probability
A coin is tossed 5 times. What is the probability of no two consecutive tosses having the same result?
Title: Re: Methods 1/2 Question.
Post by: dankfrank420 on July 13, 2014, 01:20:58 pm: Quote from: idontknow2298 on July 13, 2014, 12:13:42 am
Please help me with this question, I'm struggling a lot with probability
A coin is tossed 5 times. What is the probability of no two consecutive tosses having the same result?

The only two situations where this can occur is HTHTH and THTHT

2^5 = 32 possible outcomes

2/32 = 1/16
Title: Re: Methods 1/2 Question.
Post by: Damo23 on July 17, 2014, 06:11:26 pm: Hey I'm having a bit of trouble putting these into a matrix could someone show me what it would look like in a matrix I'm not sure.

Regard Damo :)
Title: Re: Methods 1/2 Question.
Post by: keltingmeith on July 17, 2014, 06:57:17 pm: If you have two simultaneous equations of the form:
$ax + by = e \\ cx + dy = f$
Then you can place them into a matrix like so:
$ \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{cc} e \\ f \end{array} \right] $

So, with your question, they've given us a pair of simultaneous equations - but they're not in the form I gave above. So, we quickly transpose them to get them into that form:

$ d=6000+2.5n \implies d-2.5n = 6000 \\ d=4.5n - 8000 \implies -d + 4.5n = 8000 $

Then, we can put them into the matrix:
$ \left[ \begin{array}{cc} 1 & -2.5 \\ -1 & 4.5 \end{array} \right] \left[ \begin{array}{c} d \\ n \end{array} \right] = \left[ \begin{array}{cc} 6000 \\ 8000 \end{array} \right] $

Now, we just use the inverse of the first matrix to solve for d and n:

$ \left[ \begin{array}{c} d \\ n \end{array} \right] = \left[ \begin{array}{cc} 1 & -2.5 \\ -1 & 4.5 \end{array} \right]^{-1} \left[ \begin{array}{cc} 6000 \\ 8000 \end{array} \right] $

Then, just chuck that into your CAS, and you'll get a matrix of form
$ \left[ \begin{array}{c} d \\ n \end{array} \right] $
Where the top number is the answer for d (the manufacturing cost/revenue), and the bottom number is the answer for n (the amount of components).
Title: Re: Methods 1/2 Question.
Post by: Damo23 on July 17, 2014, 07:01:44 pm: Thanks but why is it 4000 instead of 8000?
Title: Re: Methods 1/2 Question.
Post by: keltingmeith on July 17, 2014, 07:03:25 pm: Because I transferred it in wrong, hahah. Should be 8000
Title: Re: Methods 1/2 Question.
Post by: Damo23 on July 17, 2014, 07:04:39 pm: Oh ok haha, thanks for the help, super helpful :)
Title: Re: Methods 1/2 Question.
Post by: M-D on August 10, 2014, 08:38:36 pm: Just on the topic of inverses, if i have the function $f(x)=4^x$ , yhe textbook gives its inverse as 1/2 log₂x. How have they simplified log₄x to 1/2 log₂x? Thanks in advance.
Title: Re: Methods 1/2 Question.
Post by: keltingmeith on August 10, 2014, 08:46:31 pm: Quote from: M-D on August 10, 2014, 08:38:36 pm
Just on the topic of inverses, if i have the function $f(x)=4^x$ , yhe textbook gives its inverse as 1/2 log₂x. How have they simplified log₄x to 1/2 log₂x? Thanks in advance.

They haven't simplified log₄(x) - they've taken a different log entirely. You're still right, don't worry, but often we like to write logs in the form of base 2, 10 or e for various reasons.

As to how they did it, $f(x)=4^x = (2^2)^x = 2^{2x}$ . Then, you just do the inverse as normal:

$y=2^{2x} \\ \therefore x=2^{2y} \\ \log_2(x) = 2y \\ y = \frac{1}{2}\log_2(x) \\ \therefore f^{-1}(x) = \frac{1}{2}\log_2(x)$
Title: Re: Methods 1/2 Question.
Post by: M-D on August 17, 2014, 07:46:46 pm: I've got another question now on exponentials. It's: solve for x: $2^x+2^{-x+1}=3$ . Please show working out. Thanks
Title: Re: Methods 1/2 Question.
Post by: IndefatigableLover on August 17, 2014, 08:21:04 pm: Quote from: M-D on August 17, 2014, 07:46:46 pm
I've got another question now on exponentials. It's: solve for x: $2^x+2^{-x+1}=3$ . Please show working out. Thanks
$2^x+2^{-x+1}-3=0$ (Multiply both sides by $2^x$ )

$\therefore 2^{2x}-3\times 2^x+2=0$ (Let $2^x=y$ )

$\therefore y^2-3y+2=0$

$\therefore y=1,2$

Sub back what 'y' equals and you'll find that your 'x' values equal 0 and 1 :)
Title: Re: Methods 1/2 Question.
Post by: M-D on August 17, 2014, 08:37:10 pm: Thanks IndefatigableLover. :)
Title: Re: Methods 1/2 Question.
Post by: Damo23 on September 09, 2014, 04:57:46 pm: Anyone here know how to do this
Find a and k such that the graph of y = a10^kx passes through the points (2, 6) and
(5, 20).

And is this like group like where all the unit 1/2 questions methods r asked or is there a bigger group?
I just was wondering :) ;)
Title: Re: Methods 1/2 Question.
Post by: dankfrank420 on September 09, 2014, 05:20:49 pm: Set up simultaneous equations (sub your x and y values in).

Solve from there.
Title: Re: Methods 1/2 Question.
Post by: Damo23 on September 09, 2014, 05:28:44 pm: Quote from: dankfrank420 on September 09, 2014, 05:20:49 pm
Set up simultaneous equations (sub your x and y values in).

Solve from there.

Yeh I've tried that already but I can't get past it