Author Topic: Silly Questions Thread (Read 21115 times) Tweet Share

Aqualim · « **on:** December 08, 2009, 09:24:06 pm »

Consider the function $f$ with the rule $f(x)=\frac{1}{2}a^{2}+ax-x^{2}$ . Sketch the graph of this function for an appropraite domain. State the domain and range of $f$ .

Ok, so I've worked out that when $x=0,~~y=\frac{1}{2}a^{2}$ . Therefore meaning my Domain = $(0,Letter~of~choice)$ .

Plus I know that my range must be from $\frac{1}{2}a^{2}$ , but how would I work out my other figure for my range? Would it be to derive the original equation and let $y=0$ ?

/0 · « **Reply #1 on:** December 08, 2009, 09:29:05 pm »

Are you only graphing only the positive half of the graph? Why not the negative half?

$f'(x) = a-2x$

Solving $f'(x) = 0$ , $x = \frac{a}{2}$ . Since the coefficient of $x^2$ is negative, this is an inverted parabola, so it will have a maximum at $x = \frac{a}{2} \Rightarrow y = \frac{3a^2}{4}$

So the range is $(-\infty, \frac{3a^2}{4}]$

Aqualim · « **Reply #2 on:** December 08, 2009, 09:35:11 pm »

This is because it is a part of an application question which requires to know the area of a box. More than likely meaning it only includes positive numbers.

I've linked the question below.

Aqualim · « **Reply #3 on:** December 08, 2009, 09:51:26 pm »

What I came up with was that the derivative of $\frac{1}{2}(a^{2}+ax-x^{2})$ was $\frac{a}{2}-x$ (correct me if i'm wrong, seeing as the CAS calc said it was only $-x$ )

Then when I made $\frac{a}{2}-x = 0$ and got $\frac{a}{2}$ , then when I sub that back into the original I get something like; $\frac{5a^2}{8}$

That seem right?

Meaning that would make the Range = $(\frac{a^2}{2},\frac{5a^2}{8}]$

TrueTears · « **Reply #4 on:** December 08, 2009, 09:55:03 pm »

Quote from: /0 on December 08, 2009, 09:29:05 pm

Are you only graphing only the positive half of the graph? Why not the negative half?

$f'(x) = a-2x$

Solving $f'(x) = 0$ , $x = \frac{a}{2}$ . Since the coefficient of $x^2$ is negative, this is an inverted parabola, so it will have a maximum at $x = \frac{a}{2} \Rightarrow y = \frac{3a^2}{4}$

So the range is $(-\infty, \frac{3a^2}{4}]$

I'd do what /0 did here, just work out the range of the graph and sketch it.

/0 · « **Reply #5 on:** December 08, 2009, 09:58:34 pm »

Quote from: Aqualim on December 08, 2009, 09:35:11 pm

This is because it is a part of an application question which requires to know the area of a box. More than likely meaning it only includes positive numbers.

I've linked the question below.

So... is the equation $f(x) = \frac{1}{2}(a^2+ax-x^2)$ not $f(x) = \frac{1}{2}a^2+ax-x^2$ ?

In that case, all seems right.

Still confused about what 'Area' this represents though.

Aqualim · « **Reply #6 on:** December 08, 2009, 10:02:32 pm »

Quote from: /0 on December 08, 2009, 09:58:34 pm

Quote from: Aqualim on December 08, 2009, 09:35:11 pm
This is because it is a part of an application question which requires to know the area of a box. More than likely meaning it only includes positive numbers.

I've linked the question below.

So... is the equation $f(x) = \frac{1}{2}(a^2+ax-x^2)$ not $f(x) = \frac{1}{2}a^2+ax-x^2$ ?

In that case, all seems right.

Still confused about what 'Area' this represents though.

Equation is $f(x) = \frac{1}{2}(a^2+ax-x^2)$
and it represents the 'BARS' Area

Ilovemathsmeth · « **Reply #7 on:** December 09, 2009, 12:24:53 pm »

Exciting question, I don't remember it from Essentials or are there extra qs in the Essentials CAS book?

Aqualim · « **Reply #8 on:** December 09, 2009, 12:35:09 pm »

Quote from: Ilovemathsmeth on December 09, 2009, 12:24:53 pm

Exciting question, I don't remember it from Essentials or are there extra qs in the Essentials CAS book?

Essentials 1 & 2 Teacher's Edition

Ilovemathsmeth · « **Reply #9 on:** December 09, 2009, 01:35:09 pm »

I see

How did you get the teacher's edition??

Aqualim · « **Reply #10 on:** December 09, 2009, 02:01:36 pm »

Quote from: Ilovemathsmeth on December 09, 2009, 01:35:09 pm

I see

How did you get the teacher's edition??

Parents friend is one of the authors of the Essentials Methods Book (or something along those lines). Just brought the CD home one day and said it might come in handy..

Ilovemathsmeth · « **Reply #11 on:** December 09, 2009, 02:53:20 pm »

Luckyyy =D

Aqualim · « **Reply #12 on:** January 05, 2010, 09:17:25 pm »

Another Question

In the expansion of $(2a-1)^n$ , the coefficient of the second term is $-192$ . Find the Value of $n$ .

ok so this is as far as I got;

Since it's the second term, $r = 1$

$\binom{n}{1}\left ( 2a \right )^{n-1}\left ( -1 \right )^1=-192$

$n\left ( 2a \right )^{n-1}\left ( -1 \right )=-192$

$n\left ( 2 \right )^{n-1}=192$

$\frac{n2^n}{2}=192$

$n2^n=384$

Now what? Calculator couldn't even solve it :S
Yet I know purely from inserting numbers that the answer is 6

TrueTears · « **Reply #13 on:** January 05, 2010, 09:18:11 pm »

lol didn't tolga or someone post a question like this.

Well yeah, there is no elementary solution.

brightsky · « **Reply #14 on:** January 05, 2010, 09:46:09 pm »

$n(2a)^{n-1}(-1) = -192 \implies n(2a)^{n-1} = 192$ doesn't it?

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