Mathematics HL/SL/Studies Help

[itty.bitty.kitty]

oh i thought i had to use the washer method because i thought that x=a was another equation oh thats easy thanks!

okay one last question for the night

let f(x) = p-(3x/x²-q²) the asymptotes are x=1,-1 and y=2 find p and q

[Drake Glau]

Not entirely sure on how to approach it but I'll go for it anyway.

Asymptotes imply that the equation is undefined at that point which in this case would be whenever the denominator is 0

So we know that 1-q² is 0 so q=1

So now we have f(x)=p-(3x/x²-1)

Pick an x value between 1 and -1, plug it in and you get an (x,y) which will leave you with just the p to solve for.

This all makes sense in my head but might not be right because I never used the y=2 asymptote so I'd appreciate anyone who could either confirm or show the right way/more efficient way if there is one

[itty.bitty.kitty]

i swear im the only one using this. Kinda awkward.

this is kinda a continuation of the previous question.

how do you show that there are no max/mins for the graph of y=2-((3x)/(x²-1))

[Drake Glau]

Take the 2nd derivative and show that it can never equal 0.

Now that I have time and for the sake of anyone else who reads this I'll add in the derivative stuff

f(x)=2-(3x/(x²-1))=2-((3x)(x²-1)^-1

Power rule, and a chain rule. The 2 derives out so just leave it alone.

f'(x)=(-3x)(-1)(x²-1)^-2(2x)+(-3x)(x²-1)^-1

Simplifies to:

f'(x)=6x²(x²-1)^-2+(-3x)(x²-1)^-1

Now do the power rule twice, and the chain rule twice also...

f''(x)=[(6x²)(-2)(x²-1)^-3(2x)+(x²-1)^-2(12x)]+[(-3x)(-1)(x²-1)^-2(2x)+(-3)(x²-1)^-1

Yea...fractions are a lot easier to show on paper than all of this crap...but whatever...

Edited February 23, 2012 by Drake Glau

[itty.bitty.kitty]

hahaha this is all me whatever more for me

find the solution to y'= -2e^x/y given that y(0)= 4

[Lero]

hahaha this is all me whatever more for me

find the solution to y'= -2e^x/y given that y(0)= 4

This looks like a separable first order differential equation. Instead of writing y', use dy/dx.

dy/dx = -2e^(x) / y

Isolate dy with the y terms and dx with the x terms:

y dy = -2e^(x) dx

Integrate both sides;

y^(2) / 2 = -2e^(x) + c.

We have the condition that y(0) = 4, or (0,4) satisfies this equation.

4^(2) / 2 = -2e^(0) + c

8 = -2 + c

c = 10

So the solution is:

y^(2) / 2 = -2e^(x) + 10 or y^(2) = -4e^(x) + 20

[MariusIBDP]

A transformation has been applied to the graph to give the graph of

a Describe fully the transformation

For all values of x:

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

b Find the values of p and q.

A transformation has been applied to the graph to give the graph of y=-x^2 + 4x

c using your answer to part b, or otherwise, describe fully the transformation

I'm mostly struggling with finding value of p and describe the transformation using answer to part b, but I would appreciate if you gave the full answer.

Edited March 6, 2012 by MariusIBDP

[CkyBlue]

It would help if you draw out the graphs. Then you would see the transformations/stretches/compressions clearly.

1. The difference in x^2 and -x^2 is that it is reflected along the x axis.

2. The second expression in that equation is in vertex form. Since the expression for the parabola is given on the left side of the equation, you can easily find that. First you need to find the vertex of the parabola:

y=x^2 +4x

The equation to find the max/min x-value for the parabola is given by x=-b/2a ... looking at that we see b=4, a=1

x=-4/2

x=-2

Plug x=-2 into the equation and you have

y=-4

When you are given an equation in the form of (x+p)^2 + q, you know that -p is the x coordinate of the vertex and q is the y coordinate. Therefore, p=2 and y=-4

3. Draw the parent function, x^2, and draw the graph of the given function. You can make the connection that it's been flipped and the points moved. An easy way to do this is to compare the vertex of the two graphs.

[itty.bitty.kitty]

what does pi^- mean? I am doing limits and it says as x approaches pi^-.....and another problem says 2^-

(sinx/(1-cosx) as x approaches pi^- for the limit can someone help me to solve this? i have no idea how to do it....

[Drake Glau]

the superscript of - next to the value the limit goes to means that the limit is takin as X approaches that value from the LEFT (the negative side, hence the negative sign). Yes, the superscript + means from the right, or positive side.

Isn't it just 0? Plug in pi and you get 0/2 (sinpi=0 and 1-cospi=1-(-1)=2...)

[opus92fan]

If given a function, f(x) and asked to find -f(x), how would you go about drawing -f(x) if the equation of f(x) is not given, just a picture? A reference to a specific question would be question #5a on May 2008's TZ2 exam.

[LMaxwell]

If given a function, f(x) and asked to find -f(x), how would you go about drawing -f(x) if the equation of f(x) is not given, just a picture? A reference to a specific question would be question #5a on May 2008's TZ2 exam.

It'll just be a reflection of the picture. In your case, the f(x) curve will need to reflected through the x-axis to give -f(x)

[IB Smacher Glau]

I need help !!

how do i find the inverse of f(x) = 2 e^x – e^-x

first i know that we should say y=2 e^x – e^-x

and then interchange the position of the x and y

however the answer is completely different in the marks heme

any help will be be appreciated !

[Grassroot]

I have a doubt regarding calculator use in IB exam

I have Stat as my option for P3 HL

however something I see from the MS is that MS does not have a MS for using calculators to solve the problem....

I think using calculators to solve stat problems should be awarded with full marks

E.G. to solve chi-square problem using GOF test in TI84+

However, I was never confirmed by an authority about this. I wonder if my understanding of the problem is correct

[Drake Glau]

As far as i can tell, the calculator would do all the work for you. This shows IB nothing except that you can use a calculator. You shouldn't get full points for that in my opinion. The markschemes show the points awarded for all the work required to do the problem. There will probably not be ANY questions that you can solve with just the calculator. It will be there to aid you in speed and possibly intermediate steps for a math process. Not the entire question.

[Dinstruction]

Is the data booklet that says "First Examinations in 2006" current? Or is there a newer version with different reference formulas for the exam?

[Grassroot]

As far as i can tell, the calculator would do all the work for you. This shows IB nothing except that you can use a calculator. You shouldn't get full points for that in my opinion. The markschemes show the points awarded for all the work required to do the problem. There will probably not be ANY questions that you can solve with just the calculator. It will be there to aid you in speed and possibly intermediate steps for a math process. Not the entire question.

I get your point. But on the other hand marking scheme mentions to award marks to alternative methods of solving problems. I wonder if explicit description of calculator usage can award me full marks.

[itty.bitty.kitty]

This is probably really easy but how do you solve 2^2x+2-10*2^x +4=0

[The Economist]

This is probably really easy but how do you solve 2^2x+2-10*2^x +4=0

Note: I couldn't write it on math type but I let 2^x=w

[timtamboy63]

I need help !!

how do i find the inverse of f(x) = 2 e^x – e^-x

first i know that we should say y=2 e^x – e^-x

and then interchange the position of the x and y

however the answer is completely different in the marks heme

any help will be be appreciated !

I noticed no-one answered this, and the advice you were given was completely wrong. It took me a very, very long time to work this out, it was a bloody hard question. Here's my solution:

http://i.imgur.com/lfyRc.png

Sorry it's so big, I uploaded the scanned image.