Author Topic: onur369's Methods Question Thread :) (Read 30542 times) Tweet Share

onur369 · « **on:** February 03, 2011, 09:22:17 pm »

Ill definitely have multiple questions which need answering through out the year.

So Ill just kick start it. I attached two questions below.

taiga · « **Reply #1 on:** February 03, 2011, 09:32:26 pm »

First one; Simultaneous equations.

And if (3,3) is a stationary point you can also differentiate it and let it equal 0 for x=3 for another equation.

m@tty · « **Reply #2 on:** February 03, 2011, 09:39:13 pm »

13.

a)
From inspection, b=-3 and c=3. Deduced by basic transformation rules. It's a basic cubic. And the point of inflection is given by (-b, c) in this case.

Sub, (2, 0): $f(2)=0=a(2-3)^3+3=-a+3$

$\therefore a=3$

b)
g(x) is a reflection in the y-axis, so $g(x)=f(-x)=3(-x-3)^3+3=3(-(x+3))^3+3=-3(x+3)^3+3$

range is the same: [0,6]

Domain is [-4,-2]

c)

Height of 3.375 is the value of f(x), so find x such that f(x)=3.375

$3.375=3(x-3)^3+3$

$\implies \frac{0.375}{3}=(x-3)^3$

$\implies x=(0.125)^{\frac{1}{3}}+3=0.53368033+3=3.534\text{ cm}$

Now, the width is twice this distance as it is the value between f and g.

So the width is 7.067cm.

onur369 · « **Reply #3 on:** February 03, 2011, 09:48:40 pm »

Thanks a lot m@tty, +1.

Anyone know how to do the second one.

m@tty · « **Reply #4 on:** February 03, 2011, 10:00:09 pm »

$d(t)=at^2(b-t)$

Points given:
(2, 3) - $d(2)=3=4a(b-2)$ ...(1)

(5, 0) - $d(5)=0=25a(b-5)$ ...(2)

From (2) $b=5$ , (presuming $a,b\ne 0$ )

Sub that into 1: $3=4a\times 3 \implies 12a=3 \implies a=\frac{1}{4}$

So, $d(t)=\frac{1}{4}t^2(5-t)$

As $d(t), t>0$ , $t\in [0,5]$

To find the range you need to use calculus.

$d'(t)=\frac{1}{4}(10t-3t^2)$

Let $d'(t)=0=t(10-3t)$

So $t=0, \frac{10}{3}$

So max occurs at t=10/3

$d\left(\frac{10}{3}\right)=\frac{1}{4}\left(\frac{10}{3}\right)^2\left(5-\frac{10}{3}\right)=4.630\text{ km}$

So the range is $d(t)\in [0,4.630]$

And to the closest 100m the max is 4600m, and this occurs at t=3.333 hours ie. 200 minutes

onur369 · « **Reply #5 on:** February 03, 2011, 10:05:47 pm »

Legend, +1 you later again.

m@tty · « **Reply #6 on:** February 03, 2011, 10:22:41 pm »

Ah it's cool.. It's good to help everyone out.

My only requirement is that you now get 50

(seriously, ask andiio)

onur369 · « **Reply #7 on:** February 03, 2011, 10:25:58 pm »

hahahahaha, I would never get a 50. At most 45 if I study 24/7. But a 40 will be nice.

MrIraq · « **Reply #8 on:** February 04, 2011, 12:11:58 am »

Quote from: onur369 on February 03, 2011, 10:25:58 pm

hahahahaha, I would never get a 50. At most 45 if I study 24/7. But a 40 will be nice.

dream on...

onur369 · « **Reply #9 on:** February 04, 2011, 10:04:27 pm »

Quote from: MrIraq on February 04, 2011, 12:11:58 am

Quote from: onur369 on February 03, 2011, 10:25:58 pm
hahahahaha, I would never get a 50. At most 45 if I study 24/7. But a 40 will be nice.

dream on...

Stfu retard, go get a 35 in legal without knowing what a plaintiff and verdict is..

onur369 · « **Reply #10 on:** February 06, 2011, 03:47:58 pm »

Hey guys, Just got caught up with a question. I can do it but Im wondering if there is a quicker way to do it.

If the functions

and

both cross the x-axis at -1, determine the values for a and b.

NOTE: First function equals to f(x)

Greatness · « **Reply #11 on:** February 06, 2011, 03:51:36 pm »

Quote from: onur369 on February 06, 2011, 03:47:58 pm

Hey guys, Just got caught up with a question. I can do it but Im wondering if there is a quicker way to do it.

If the functions (Image removed from quote.) and (Image removed from quote.) both cross the x-axis at -1, determine the values for a and b.

NOTE: First function equals to f(x)

SUb x=-1 and y=0 into f(x) and g(x) then solve for a and b via simutaneous equations.

onur369 · « **Reply #12 on:** February 06, 2011, 04:09:32 pm »

The results I obtained were: a= -5 and b= -7. Correct?

Greatness · « **Reply #13 on:** February 06, 2011, 04:25:05 pm »

Quote from: onur369 on February 06, 2011, 04:09:32 pm

The results I obtained were: a= -5 and b= -7. Correct?

I just chucked it all into my calculator, i got a=4 and b=-16. Maybe my method is wrong? :s

xZero · « **Reply #14 on:** February 06, 2011, 04:34:55 pm »

Quote from: swarley on February 06, 2011, 04:25:05 pm

Quote from: onur369 on February 06, 2011, 04:09:32 pm
The results I obtained were: a= -5 and b= -7. Correct?
I just chucked it all into my calculator, i got a=4 and b=-16. Maybe my method is wrong? :s

your method should be correct and i got a=4 and b=-16 as well

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