Mathematics HL/SL/Studies Help

[CkyBlue]

Oh thank you...I know that Capt'n Marth helped me on a similar problem before...but I am not sure if I did it right....whenever I get weird answers I kinda freak out even though that is not good...

Don't fret if you don't obtain integer answers. Have faith in your ability to think critically. Real life problems almost never involve integers, and if you end up getting a weird decimal and have checked your work, so be it.

[Guest hellokitty818]

Sorry got a math test tomorrow....How do you do this...

The point B (a,b) is on the curve f(x)=x^2 such that B is the point which is closest to A(6,0). Calculate the value of a.

[Lero]

Sorry got a math test tomorrow....How do you do this...

The point B (a,b) is on the curve f(x)=x^2 such that B is the point which is closest to A(6,0). Calculate the value of a.

This question basically uses the distance between two points formula and differentiation to find the minimum distance.

Your graph given is f(x) = x^(2), and your co-ordinates are (a,b). That b can be replaced by a^(2) simply by substituting f(a) = a^(2).

Therefore you now have both your co-ordinates in terms of a. Using the distance between two points formula:

sqrt((a^2-0)^2+(a-6)^2) = distance. d/da = 0 gives you the minimum distance.

Differentiating and simplifying leads to 4a^(3) + 2a - 12 = 0. Using a GDC I got the answer to be a = 1.33

Edited October 25, 2011 by Hus

[Guest hellokitty818]

Sorry got a math test tomorrow....How do you do this...

The point B (a,b) is on the curve f(x)=x^2 such that B is the point which is closest to A(6,0). Calculate the value of a.

This question basically uses the distance between two points formula and differentiation to find the minimum distance.

Your graph given is f(x) = x^(2), and your co-ordinates are (a,b). That b can be replaced by a^(2) simply by substituting f(a) = a^(2).

Therefore you now have both your co-ordinates in terms of a. Using the distance between two points formula:

sqrt((a^2-0)^2+(a-6)^2) = distance. d/da = 0 gives you the minimum distance.

Differentiating and simplifying leads to 4a^(3) + 2a - 12 = 0. Using a GDC I got the answer to be a = 1.33

i got everything until 4a^(3) + 2a - 12 = 0....how does (a^2)^2 + (a-6)^2 =4a^(3) + 2a - 12 .....

doesnt it have to be a^4+a^2-12a+36?? or did i just make a sill mistake...i am sorry i just do not see it...

[Lero]

Sorry got a math test tomorrow....How do you do this...

The point B (a,b) is on the curve f(x)=x^2 such that B is the point which is closest to A(6,0). Calculate the value of a.

This question basically uses the distance between two points formula and differentiation to find the minimum distance.

Your graph given is f(x) = x^(2), and your co-ordinates are (a,b). That b can be replaced by a^(2) simply by substituting f(a) = a^(2).

Therefore you now have both your co-ordinates in terms of a. Using the distance between two points formula:

sqrt((a^2-0)^2+(a-6)^2) = distance. d/da = 0 gives you the minimum distance.

Differentiating and simplifying leads to 4a^(3) + 2a - 12 = 0. Using a GDC I got the answer to be a = 1.33

i got everything until 4a^(3) + 2a - 12 = 0....how does (a^2)^2 + (a-6)^2 =4a^(3) + 2a - 12 .....

doesnt it have to be a^4+a^2-12a+36?? or did i just make a sill mistake...i am sorry i just do not see it...

I skipped a few steps sorry.

Did you use the chain rule to differentiate the stuff inside the radical?

Once you do that it should become more clearer as you equate the derivative to zero and you should be left with 4a^(3) + 2a - 12 = 0.

Edited October 25, 2011 by Hus

[scarlettjazz]

2x^2 -6 = x^3 - 5x

(without a calculator) Thanks!

[Lero]

2x^2 -6 = x^3 - 5x

(without a calculator) Thanks!

Rearrange the equation bringing all terms to one side: x^(3) - 2x^(2) - 5x + 6 = 0.

Now it is a matter of trial and error to determine the initial zero of the function. If the question does not allow for a calculator then your best bet is trying x = -3, -2, -1, 0, 1, 2 and 3 to see if you can find a zero.

In this case x = 1 is a zero of the function, therefore a factor is (x - 1).

Using polynomial division you can determine the other factors, it is very tedious and probably too hard to show on here

Dividing (x - 1) into the function gives (x - 1)(x^(2) - x - 6). Factoring the quadratic gives:

(x - 1)(x - 3)(x + 2) = 0. Therefore x = 1, 3 and -2.

[dessskris]

just a tip, for the trial and error part you should also try +/-0.5, sometimes works!

another tip, sometimes you can guess from the constant in the equation. if it's 6, you can usually try to plug in 1, 6, 2 or 3 for the first trial and error part. doesn't always work, but most of the time it does.

Edited October 25, 2011 by Desy Glau

[Wide Eyed Wanderer]

When they haven't already given you a root, you'll have to use the trial and error method... Fortunately in IB you wont generally get equations in which you have to try more than -1, -0.5, , 0.5, 1, 2, -2..

That'll give you your first factor, after which you can factorise the rest of the equation using division. If you find that too tedious, you can use the synthesis method ( aka Ruffini's method). It will make your life a lot simpler

[aldld]

Disclaimer: I'm only starting my first year of IB and our class is nowhere near this material yet, so I may be horribly wrong. But I'll take a crack at it anyway.

(a) Define an isomorphism between two groups (G, * ) and (H, •).

Is this just asking for the definition? In this case, an isomorphism between two groups (G, * ) and (H, •) would be a bijective function f: G->H such that f(a*b) = f(a) • f(b).

(b) Let e and e' be the identity elements of groups G and H respectively.

Let f be an isomorphism between these two groups. Prove that f (e) = e'.

e*e = e

f(e*e) = f(e) = f(e) • f(e)

f(e) • f(e)^-1 = f(e) • f(e) • f(e)^-1

f(e) • f(e)^-1 = e'

e' = f(e) • e'

e' = f(e)

I don't think I see any problems with this proof, I'm just sort of teaching myself this on the go... I could try helping you but my main reason for doing this is kind of selfish I guess

© Prove that an isomorphism maps a finite cyclic group onto another

finite cyclic group.

I only have a cursory awareness of cyclic groups, but I'll definitely take a look at that when I get a bit of spare time...

[jonathan810]

Let f(x)=x² and g(x)=2(x-1)²

a: the graph of g can be obtained from the graph of 'f' using 2 transformations. give a full geometric description of each of the 2 transformations.

b) the graph of 'g' is translated by the vector (3,-2) (formatting error: the 3 is on top and the -2 is on the bottom) to give the graph of 'h'. The point (-1,1) on the graph of 'f' is translated to the point 'P' on the graph of 'h'. Find the coordinates of 'P'

Please help!!!!!

[CkyBlue]

For question A:

The parent function is a quadratic. I don't really think it's two transformations, it's one transformation and a stretch.

The stretch is a vertical stretch by a factor of 2, and there is a horizontal translation one unit to the right.

If you apply those changes to the graph of f, you will get the graph of g.

For question B: Why are there 3 graphs? I'm pretty sure one isn't relevant...Correct me if I'm wrong.

Just move the point of (-1,1) 3 units right, 2 units down. Point P is (2,-1)

Haven't touched math in a while, can anyone confirm this?

PS: Damn, these are pretty easy questions, I hope you're ready since you're exams are in Nov!

[jonathan810]

the first part is correct (I managed to get that part)

but the second part is where I get stuck, the answers say that 'P'=(3,0) and can be gathered through the co-ordinates in 'g'

[dessskris]

the first part is correct (I managed to get that part)

but the second part is where I get stuck, the answers say that 'P'=(3,0) and can be gathered through the co-ordinates in 'g'

he forgot the change of f to g first.

from f (-1,1)

to g (0, 2) due to vertical stretch x2 and horizontal translation 1 unit to the right

to h (3, 0) due to vector (3,-2) translation

[Procrastination]

Help me please?

Expand:

(2+x-4x²)²

I was thinking of saying (2+x) is just a single term and then developing it as another polynomial expression in the answear. But im not really sure, could you please solve it an explain it?

[Emmi]

It had lots of exponents in it so I wrote it out.

So (2+x-4x²)² is just (2+x-4x²)(2+x-4x²).

You distribute each term, first the 2, then the x, then the -4x².

Then I just grouped like terms together (terms with the same power) and added them to simplify.

[Procrastination]

.Thank you amy! I got a similar result to the one you are giving me here with another method. However, I beg to differ in the "+15x²". I got "-15x²". Hum, kinda weird. Anyways, thank you. Thumbs up!

EDIT:

I revised your answear and it is indeed "-15x²". You just confused the signs. -8x² -8x²+x² is -15x².

Edited October 30, 2011 by Procrastination

[Emmi]

You're right, it should be -15x² Just a little mistake, sorry. I think I made it positive on accident when I was adding the x².

[Procrastination]

Don't worry. It really cleared up some things for me.

Edited October 30, 2011 by Procrastination

[Drake Glau]

2x^2 -6 = x^3 - 5x

(without a calculator) Thanks!

Rearrange the equation bringing all terms to one side: x^(3) - 2x^(2) - 5x + 6 = 0.

Now it is a matter of trial and error to determine the initial zero of the function. If the question does not allow for a calculator then your best bet is trying x = -3, -2, -1, 0, 1, 2 and 3 to see if you can find a zero.

In this case x = 1 is a zero of the function, therefore a factor is (x - 1).

Using polynomial division you can determine the other factors, it is very tedious and probably too hard to show on here

Dividing (x - 1) into the function gives (x - 1)(x^(2) - x - 6). Factoring the quadratic gives:

(x - 1)(x - 3)(x + 2) = 0. Therefore x = 1, 3 and -2.

We learned something in calc last week to help with this kind of stuff called Newton's Method. Basically allows you to estimate the root using the derivative of the function. Not sure if the IB curriculum teaches this or not but it might be a good thing to learn in case those initial integers aren't the zeros or even close enough to guess.

http://en.wikipedia.org/wiki/Newton's_method

Basically as you keep doing it over and over, the x value keeps approaching the root.