2 unit maths revision (1 Viewer)

[jessica2011hsc]

my school is behind in this topic, so im trying to accelerate. that's why i asked this question, cause im not completely sure how to do it with confidence

 

[SpiralFlex]

my school is behind in this topic, so im trying to accelerate. that's why i asked this question, cause im not completely sure how to do it with confidence

Curve is increasing. (Curve going down at a particular range)
Curve is decreasing. (Curve going up at a particular range)
Curve stationary. (Curve not moving at all at particular points)

Step 1: Differentiate.

Step 2: For stationary points make

Step 3: Solve.

Step 4: Substitute back into ORIGINAL equation.

Step 5: You have your stationary points.

 

[jessica2011hsc]

i appreciate your help

 

[SpiralFlex]

No worries, I have nothing better to do on a Saturday night anyway.

 

[hscishard]

P(x)=ax^3+bx^2+cx+d P(0) = 5 = d
P'(x)=3ax^2+2bx+c
0 = 3ax^2 + 2bx + c for stat points.
0 = 3a + 2b + c 0 = 12a - 4b + c
12a - 3a - 4b -2 b + c - c = 0
9a - 6b = 0
3a - 2b = 0
P(1) = 12 P(-2) = 15
Hence, a + b + c + d = 12 -8a + 4b - 2c + d = 15
a + 8a + b - 4b + c + 2c + d - d = 12 - 15
9a - 3b + 3c = -3
3a - b + c = -1
3a - 3a + b + 2b + c - c = 0 + 1
3b = 1
b = 1/3
9a - 5(1/3) = 0
9a = 2.5
a = 5/27
3(5/27) - 1/3 + c = -1
c = -1 - 15/27 + 1/3
c = -11/9
a + b + c + d = 12
5/27 + 1/3 - 11/9 + d = 12
d = 12 - 5/27 -1/3 + 11/9
This is wrong... lol

lol: know yu cant said 2, unit mths is piss esy

Lmao

 

[AAEldar]

Stationary when





Substitute back,

I haven't worked it out, but from the line:



Doesn't ?

 

[SpiralFlex]

Correct, thank you for correction.

 

[HSCAREA]

Make y the subject,find intersection points and graph it. Then you just do it

this i'm am finding difficulty. please provide some form of help : [

 

[Riproot]

lol: know yu cant said 2, unit mths is piss esy

Lmao

What is wrong with your language?

 

[hscishard]

(x+2)^2 = 4x-x^2
Then solve

 

[HSCAREA]

(x+2)^2 = 4x-x^2
Then solve

x^2 +4x + 4 = 4x - x^2

4x = 4x - 4

x = x - 1 the fuck?


__________________________

(x+2)(x+2) = x^2 + 4x + 4

therefore x^2 + 4x + 4 = 4x - x^2
2x^2 + 4 = 0

therefore 2x^2 = -4

x^2 = -2

x = sqroot -2

= no real roots the fuck?

 

[Bored Of Fail1]

you made mistake noob ^

x^2 +4 =-x^2

but this gives non real solns

 

[Bored Of Fail1]

(x+2)(x+2) = x^2 + 4x + 4

therefore x = 4x - x^2 - 2

lol thats the worst display of algebra I have ever seen

ab=0

means a=0 or b=0

that only works when a product equals zero!! : (

EDIT: but even if you were trying x+2 = x^2 + 4x+4 that would give x= x^2 +4x + 2

lolz so many mistakes

 

[slyhunter]

.....

 

[benji_10]

The solution is in the complex field, which is 4U. I thought this was 2U revision?

 

[hscishard]

Sorry, I didn't see the function correctly.
(x+2)^2 = 4-x^2
Since I've caused you some confusion, i'll post the solution

x^2 + 4x +4 = 4-x^2
2x^2 +4x = 0
x=0 and -2. Check by subbing it back into each function to make sure they yield the same value. In this case they do.

Now it's
You can calculate the first one quite easily. It's just 1/4 of the area of a circle. The radius is 2, so pi(2)^2 / 4 = pi
Now it's

The integral isn't hard to find, i'll assume you know how to do it.
The answer should be pi-2

 

[deterministic]

Question: http://imgur.com/SZkDD

um hii guys, I'm stuck on this question and don't know what the intercept of the 2 curves are. thankss

To all above attempting this question, learn to read properly.

You are finding points of intersection of:
and

So

 

[hscishard]

To all above attempting this question, learn to read properly.

Ha, no it was just me. I think they assumed what I was saying was correct
Talk about assuming people are correct...

 

[lalala369]

differentiate <a href="http://www.codecogs.com/eqnedit.php?latex=\displaystyle x\sqrt{x@plus;1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\displaystyle x\sqrt{x+1}" title="\displaystyle x\sqrt{x+1}" /></a>

 

[ohexploitable]

differentiate <a href="http://www.codecogs.com/eqnedit.php?latex=\displaystyle x\sqrt{x@plus;1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\displaystyle x\sqrt{x+1}" title="\displaystyle x\sqrt{x+1}" /></a>