Maths Methods 3/4 Help Thread 2011

[vgardiy]

This question is from C*E 2011 so dont read it if you plan on doing the exam.

Find the tangent to both curves: y=x^2 - x and y=x^2 + 2x

I was wondering if there was an easy way to do this.

[BlueSky_3]

@vgardiy: Do you happen to have an online copy of any of the C*E methods exams that you could perhaps pm me?
My school didn't seem to have any from that company at all 

[luken93]

Is legit notation for a condition?

Perhaps 1 + 2n, n c R ? haha just a suggestion

Don't you mean n c Z?

Eh that haha

[brightsky]

let a and b be where the required line touches the first and second parabola respectively.
y - a^2 + a = (2a-1)(x-a) ==> y = (2a-1)x - a^2 - 2a
y - b^2 - 2b = (2b+2)(x-b) ==> y = (2b+2)x - b^2
so we want both lines to be equal:
(2a - 1)x - a^2 - 2a =  (2b + 2)x - b^2
2a - 1 = 2b + 2 ==> b = (2a-3)/2
a^2 + 2a = b^2
so a^2 + 2a = (2a-3)^2/4
4a^2 + 8a = 4a^2 - 12a + 9
20a = 9
a = 9/20
b = -21/20
so the required line is y = -x/10 - 441/400

[BlueSky_3]

Really silly question but how does -(sqrt(2))^3 +4(sqrt(2)) = 2(sqrt(2)) ?

[nacho]

insight is useless for methods
agree / disagree?

[vgardiy]

let a and b be where the required line touches the first and second parabola respectively.
y - a^2 + a = (2a-1)(x-a) ==> y = (2a-1)x - a^2 - 2a
y - b^2 - 2b = (2b+2)(x-b) ==> y = (2b+2)x - b^2
so we want both lines to be equal:
(2a - 1)x - a^2 - 2a =  (2b + 2)x - b^2
2a - 1 = 2b + 2 ==> b = (2a-3)/2
a^2 + 2a = b^2
so a^2 + 2a = (2a-3)^2/4
4a^2 + 8a = 4a^2 - 12a + 9
20a = 9
a = 9/20
b = -21/20
so the required line is y = -x/10 - 441/400

I did that method, but i got y= 1/2 x - 9/16 :S

[paulsterio]

insight is useless for methods
agree / disagree?

it's not THAT bad, but obviously it doesn't stand up against Puffy Publications

[brightsky]

Really silly question but how does -(sqrt(2))^3 +4(sqrt(2)) = 2(sqrt(2)) ?

yes.

[nacho]

insight is useless for methods
agree / disagree?

it's not THAT bad, but obviously it doesn't stand up against Puffy Publications

puffy publications better look out for nulkit.g nachokit.g publications
but i found the insight 2 exam to  be much like an exam 1

[nacho]

If f( x) = h(x(x+2)) , then f’(x) =  is equal to ?
my workings:
I got h'(x^2+2x) + h(x^2 + 2x)'
= h'(x^2+2x) + h(2x + 1)
= doesnt fit any options


also.. how do you guys go about sketching hybrid graphs like
f(x) =        {  1/1000(x-20)               20<x<40
f(x)=         { 1/4000(x+40)                   40<x<120

its not just the hybridness, but the crazy dilations.. i can never get a good scale

[BlueSky_3]

Ok based on my interpretation is the question f(x) = h(x^2+2x) ?

If so, then f'(x) = (2x+2) x h'(x^2 + 2x) , using chain rule

[luffy]

f(x ) = h(x^2 + 2x).

Let y = f(x ) = h(u), where u = x^2 + 2x.

f'(x) = dy/dx = dy/du x du/dx

= h'(u) x (2x + 2)

= (2x + 2) x h'(x^2 + 2x)

[nacho]

Extended response chapter 1 pg 54 q6a
how do you figure out the inner x intercepts? The outer two are fairly obvious.. But the inner two?

[paulsterio]

Extended response chapter 1 pg 54 q6a
how do you figure out the inner x intercepts? The outer two are fairly obvious.. But the inner two?

What book?