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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Hielly on November 28, 2009, 11:46:47 pm

Title: Help
Post by: Hielly on November 28, 2009, 11:46:47 pm: Need help on 11(d)(f)
(http://i150.photobucket.com/albums/s89/helenn--/math-2.jpg)

thanks
Title: Re: Help
Post by: TrueTears on November 28, 2009, 11:48:11 pm: Best way is to sketch the graph and find the x values.

For d) you should get $x < -1$ or $x > \frac{2}{3}$

for f) try this:

$3x^2+x-2 \le -2$

$3x^2+x \le 0$

again sketch graph.
Title: Re: Help
Post by: Hielly on November 28, 2009, 11:50:54 pm: Quote from: TrueTears on November 28, 2009, 11:48:11 pm
Best way is to sketch the graph and find the x values.

For d) you should get $x < -1$ or $x > \frac{2}{3}$

for f) try this:

$3x^2+x-2 \le -2$

$3x^2+x \le 0$

again sketch graph.
so i did (3x-2)(x+1) greater than 0

so x=2/3 and x=-1

how do you know if its greater or less than? you can tell by drawing a graph and looking at it right?
Title: Re: Help
Post by: TrueTears on November 28, 2009, 11:51:29 pm: A more algebraic way (albeit longer!) would be this:

for d)

$(x+1)(3x-2) > 0$

Case 1:

$x+1 > 0$ and $3x-2 >0$

$x > -1$ and $x > \frac{2}{3}$

The 'general' solution here is $x > \frac{2}{3}$

Case 2:

$x+1<0$ and $3x-2 < 0$

$x< -1$ and $x < \frac{2}{3}$

The 'general' solution here is $x<-1$

Combine the 2 cases and you get $x<-1$ or $x > \frac{2}{3}$

Notice the use of "and" and "or" when splitting cases, it is crucial that you don't get mixed up with these 2 important words.
Title: Re: Help
Post by: TrueTears on November 28, 2009, 11:54:24 pm: Quote from: Hielly on November 28, 2009, 11:50:54 pm
Quote from: TrueTears on November 28, 2009, 11:48:11 pm
Best way is to sketch the graph and find the x values.

For d) you should get $x < -1$ or $x > \frac{2}{3}$

for f) try this:

$3x^2+x-2 \le -2$

$3x^2+x \le 0$

again sketch graph.
so i did (3x-2)(x+1) greater than 0

so x=2/3 and x=-1

how do you know if its greater or less than? you can tell by drawing a graph and looking at it right?

I assume you are talking about d)

If so yes it is just the x intercepts. Then you must find the values of x for which the graph is strictly larger than 0. This is easily seen from the graph.
Title: Re: Help
Post by: Hielly on November 29, 2009, 12:01:20 am: Quote from: TrueTears on November 28, 2009, 11:51:29 pm
A more algebraic way (albeit longer!) would be this:

for d)

$(x+1)(3x-2) > 0$

Case 1:

$x+1 > 0$ and $3x-2 >0$

$x > -1$ and $x > \frac{2}{3}$

The 'general' solution here is $x > \frac{2}{3}$

Case 2:

$x+1<0$ and $3x-2 < 0$

$x< -1$ and $x < \frac{2}{3}$

The 'general' solution here is $x<-1$

Combine the 2 cases and you get $x<-1$ or $x > \frac{2}{3}$

Notice the use of "and" and "or" when splitting cases, it is crucial that you don't get mixed up with these 2 important words.
Interesting, thanks!
Title: Re: Help
Post by: cipherpol on November 29, 2009, 11:10:56 am: Quote from: TrueTears on November 28, 2009, 11:51:29 pm
A more algebraic way (albeit longer!) would be this:

for d)

$(x+1)(3x-2) > 0$

Case 1:

$x+1 > 0$ and $3x-2 >0$

$x > -1$ and $x > \frac{2}{3}$

The 'general' solution here is $x > \frac{2}{3}$

Case 2:

$x+1<0$ and $3x-2 < 0$

$x< -1$ and $x < \frac{2}{3}$

The 'general' solution here is $x<-1$

Combine the 2 cases and you get $x<-1$ or $x > \frac{2}{3}$

Notice the use of "and" and "or" when splitting cases, it is crucial that you don't get mixed up with these 2 important words.
This may be stupid, but from the two solutions for x, how do you pick out the general solution?
Title: Re: Help
Post by: samuch on November 29, 2009, 11:44:09 am: Quote from: cipherpol on November 29, 2009, 11:10:56 am
Quote from: TrueTears on November 28, 2009, 11:51:29 pm
A more algebraic way (albeit longer!) would be this:

for d)

$(x+1)(3x-2) > 0$

Case 1:

$x+1 > 0$ and $3x-2 >0$

$x > -1$ and $x > \frac{2}{3}$

The 'general' solution here is $x > \frac{2}{3}$

Case 2:

$x+1<0$ and $3x-2 < 0$

$x< -1$ and $x < \frac{2}{3}$

The 'general' solution here is $x<-1$

Combine the 2 cases and you get $x<-1$ or $x > \frac{2}{3}$

Notice the use of "and" and "or" when splitting cases, it is crucial that you don't get mixed up with these 2 important words.
This may be stupid, but from the two solutions for x, how do you pick out the general solution?
ill take a guess :) since with case 1 the signs being used at the start are 'greater than' zero then you pick the answer that is greater than zero and for case 2 since the sign at the start being used is the 'less than' zero sign then you pick the answer that is less than zero

Edit: remember that i may be completely wrong tho!
Title: Re: Help
Post by: cipherpol on November 29, 2009, 11:47:53 am: Quote from: samuch on November 29, 2009, 11:44:09 am
Quote from: cipherpol on November 29, 2009, 11:10:56 am
Quote from: TrueTears on November 28, 2009, 11:51:29 pm
A more algebraic way (albeit longer!) would be this:

for d)

$(x+1)(3x-2) > 0$

Case 1:

$x+1 > 0$ and $3x-2 >0$

$x > -1$ and $x > \frac{2}{3}$

The 'general' solution here is $x > \frac{2}{3}$

Case 2:

$x+1<0$ and $3x-2 < 0$

$x< -1$ and $x < \frac{2}{3}$

The 'general' solution here is $x<-1$

Combine the 2 cases and you get $x<-1$ or $x > \frac{2}{3}$

Notice the use of "and" and "or" when splitting cases, it is crucial that you don't get mixed up with these 2 important words.
This may be stupid, but from the two solutions for x, how do you pick out the general solution?
ill take a guess :) since with case 1 the signs being used at the start are 'greater than' zero then you pick the answer that is greater than zero and for case 2 since the sign at the start being used is the 'less than' zero sign then you pick the answer that is less than zero

Edit: remember that i may be completely wrong tho!
haha, I'll trust that you're correct. Thanks for the help!
Title: Re: Help
Post by: Ilovemathsmeth on November 29, 2009, 01:48:46 pm: I always found it easier to graph the function. For part (f), I'd draw in the line y = -2. I'd find where the function equals -2. Then judging from the graph, you can decide where the function is greater than -2 and less than -2.

Same applies for f(x) > 0, find where it equals zero then from the graph you can easily tell where it is greater.

How do I write in that maths font? I really like it!
Title: Re: Help
Post by: Gloamglozer on November 29, 2009, 02:27:28 pm: Quote from: Ilovemathsmeth on November 29, 2009, 01:48:46 pm
How do I write in that maths font? I really like it!

You'll need to learn LaTex. It's like BB coding but google "LaTex" and you'll find some good resources.
Title: Re: Help
Post by: TrueTears on November 29, 2009, 03:29:29 pm: Quote from: cipherpol on November 29, 2009, 11:10:56 am
Quote from: TrueTears on November 28, 2009, 11:51:29 pm
A more algebraic way (albeit longer!) would be this:

for d)

$(x+1)(3x-2) > 0$

Case 1:

$x+1 > 0$ and $3x-2 >0$

$x > -1$ and $x > \frac{2}{3}$

The 'general' solution here is $x > \frac{2}{3}$

Case 2:

$x+1<0$ and $3x-2 < 0$

$x< -1$ and $x < \frac{2}{3}$

The 'general' solution here is $x<-1$

Combine the 2 cases and you get $x<-1$ or $x > \frac{2}{3}$

Notice the use of "and" and "or" when splitting cases, it is crucial that you don't get mixed up with these 2 important words.
This may be stupid, but from the two solutions for x, how do you pick out the general solution?
For case 1, anything larger than 2/3 is larger than -1. so x>2/3 is more general

For case 2, apply same logic.
Title: Re: Help
Post by: TrueTears on November 29, 2009, 03:34:21 pm: Quote from: samuch on November 29, 2009, 11:44:09 am
Quote from: cipherpol on November 29, 2009, 11:10:56 am
Quote from: TrueTears on November 28, 2009, 11:51:29 pm
A more algebraic way (albeit longer!) would be this:

for d)

$(x+1)(3x-2) > 0$

Case 1:

$x+1 > 0$ and $3x-2 >0$

$x > -1$ and $x > \frac{2}{3}$

The 'general' solution here is $x > \frac{2}{3}$

Case 2:

$x+1<0$ and $3x-2 < 0$

$x< -1$ and $x < \frac{2}{3}$

The 'general' solution here is $x<-1$

Combine the 2 cases and you get $x<-1$ or $x > \frac{2}{3}$

Notice the use of "and" and "or" when splitting cases, it is crucial that you don't get mixed up with these 2 important words.
This may be stupid, but from the two solutions for x, how do you pick out the general solution?
ill take a guess :) since with case 1 the signs being used at the start are 'greater than' zero then you pick the answer that is greater than zero and for case 2 since the sign at the start being used is the 'less than' zero sign then you pick the answer that is less than zero

Edit: remember that i may be completely wrong tho!
Close guess but no cigar :P

What if we had (x+1)(3x-2) < 0

Case 1

x+1<0 and 3x-2>0

x<-1 and x > 2/3

Here there is NO general solution. (So you wouldn't pick the x<-1)

Case 2

x+1>0 and 3x-2<0

x>-1 and x<2/3

Now this case has the actual solution: -1<x<2/3

:)

This algebraic method is simply just to stimulate further thinking, graphing is the easiest way to go.
Title: Re: Help
Post by: samuch on November 29, 2009, 04:25:59 pm: ^ oh.. thanks for the correction :)