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The Math Help Thread

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cpp_is_king said:
It's possible I'm being dense, but I have no idea what your question is. Are you trying to solve a quadratic equation? What does A*C, A+C = B mean?
Click to expand...

I apologize, I didn't do the best at explaining it.

I don't need the answer to the equation, I need the formula that I can put into my TI-84 calculator. Under the "Y=" key, with the two line formula it will tell me what my two variables are that combine to be B in the equation. Within that window, you are shown lines "Y1=" through "Y9=". The formula my professor used was something like in line "Y1 = X + Y" and line "Y2 = B (from A * C in the equation)".

Example: 2t^2 + 11t - 63 = 0 ------ ax^2 + bx + c = 0

>To get what I want on my TI-84
>Hit "Y="
>Input "11" in line "Y2"
>hit 2ND and Table
>Table shows 3 rows
>the table tells me what A*C is and A+B are
>I take those two numbers and split the middle term
>2t^2 + 2t - 26t - 63 -- (incorrect numbers, just an example)
>factor
 
Sorry Gaf, I need help with a proof.

The question is as follows:

Suppose f: A -> B and g: B -> A.
g * f = i_A (the composition of g and f is the identity function on set A).

Prove that if g is injective, then f is surjective.

Proof:

Assume g is injective. Then we wish to prove f is surjective.
Observe that both f and g are injective functions. Then because the composition is the identity function, this must mean that the cardinalities of set A and B must be the same. Otherwise, it is impossible for the two to be injective and have the composition yield the identity for all arbitrary elements "a" in set A.

Then because the cardinalities are the same, we can infer that for each element in B, there is a pre-image in A under the function f.

Q.E.D.

That's what I have, but it's not even close to the solution.
 
Kieli said:
Sorry Gaf, I need help with a proof.

The question is as follows:

Suppose f: A -> B and g: B -> A.
g * f = i_A (the composition of g and f is the identity function on set A).

Prove that if g is injective, then f is surjective.

Proof:

Assume g is injective. Then we wish to prove f is surjective.
Observe that both f and g are injective functions. Then because the composition is the identity function, this must mean that the cardinalities of set A and B must be the same. Otherwise, it is impossible for the two to be injective and have the composition yield the identity for all arbitrary elements "a" in set A.

Then because the cardinalities are the same, we can infer that for each element in B, there is a pre-image in A under the function f.

Q.E.D.

That's what I have, but it's not even close to the solution.
Click to expand...
If you're trying to show that f is surjective, why not pick some arbitrary b in B and show that there must be some a in A so that f(a)=b? Based on what you're given, there's really only one way to pick such an a, i.e. let a=g(b). Then you have that (g(f(a))=a, since g*f is the identity on A. Now you just need to use injectivity to complete the proof.
 
An update to my earlier issue if anyone ever needs help finding the numbers to factor your rational equation out.

disclaimer: I'm using a TI-84 Plus Silver Edition

>Get your equation into ax^2 + bx + c = 0 form.
>Press Y= on your calculator.
>input into line Y1 = (A*C)/X <---- A*C is the product of those two variables (Ex. 5*5, input 10/X)
>Input for line Y2 = X + Y1
>hit 2ND and GRAPH
>look in the 3rd row for the number that equals B (ex 10)
>you can now factor into: ax^2 + 1x + 2x + c
 
Hey guys, I'm having trouble with a problem and could use some help

so the problem goes like this;

(x-3)(x+5)=18...so what I did was foil and got x^2+5x-3x-15=18. I combined like terms, subtracted 18 from both sides, set the problem equal to zero and got...x^2+2x-33=0. Following this I substitued a, b and c into the quadratic equation getting (-2 +/-&#8730;136)/2. Extracting a 2 from the &#8730;136, the current problem becomes (-2+/-2&#8730;34)/2. I then factored a 2 out of the numerator and got (2(-1 +/- 1)&#8730;34)/2. The 2's cancel one another and you then have -1 +/- 1 &#8730;34, giving an aswer of -2&#8730;34 and &#8730;34. To check it, I insert them back into the original problem but they don't equal 18. I've been trying to work this problem multiple times but nothing is working. I'd appreciate any and all help.

Thank you
 
FortuneFaded said:
This is where you fucked up.
Click to expand...

Lol help a brother out! Oh fuck, wait, hold up a minute! Is it because the 1 is attached to the radical that I can't combine it with the 1 preceding the +/-. If that's the reason, I'm going to slap 50 people before this day is done! It's crazy that I only noticed that after you told (cryptically) me what was wrong regardless of me working through the problem like 10 times
 
Allonym said:
Lol help a brother out! Oh fuck, wait, hold up a minute! Is it because the 1 is attached to the radical that I can't combine it with the 1 preceding the +/-. If that's the reason, I'm going to slap 50 people before this day is done! It's crazy that I only noticed that after you told (cryptically) me what was wrong regardless of me working through the problem like 10 times
Click to expand...

Yes, if you use -1 + &#8730;34, -1 - &#8730;34 you end up with (-4 + &#8730;34)(4 - &#8730;34) which gives the correct answer.
 
There's another solution that makes zero sense to me:

Suppose that
(n+2)/[n*(2n+3)] < &#949;

We know that (n+2)/[n*(2n+3)] <= (2n+3)/[n*(2n+3)] = 1/n

If we set N = ceiling(1/n), then for all n > N, (2n+3)/[n*(2n+3)] < &#949;.

That is PRECISELY the part I don't get. How can we conclude (2n+3)/[n*(2n+3)] = 1/n is smaller than epsilon? Our assumption is that (n+2)/[n*(2n+3)] < &#949;. All we know is that (2n+3)/[n*(2n+3)] is bigger than (n+2)/[n*(2n+3)]. We have no way of knowing that the (2n+3)/[n*(2n+3)] is also smaller than epsilon, much less use it to find some N which can be applied to find the limit of (n+2)/[n*(2n+3)].

Edit: Oh, is it the following? If f(x) < epsilon for some x > N, then given g(x) < f(x) for all x, then g(x) < epsilon?

Here, g(x) is (n+2)/[n*(2n+3)], and we elect to prove the limit of the easier function (2n+3)/[n*(2n+3)].
 
Kieli said:
There's another solution that makes zero sense to me:

Suppose that
(n+2)/[n*(2n+3)] < &#949;

We know that (n+2)/[n*(2n+3)] <= (2n+3)/[n*(2n+3)] = 1/n

If we set N = ceiling(1/n), then for all n > N, (2n+3)/[n*(2n+3)] < &#949;.

That is PRECISELY the part I don't get. How can we conclude (2n+3)/[n*(2n+3)] = 1/n is smaller than epsilon? Our assumption is that (n+2)/[n*(2n+3)] < &#949;. All we know is that (2n+3)/[n*(2n+3)] is bigger than (n+2)/[n*(2n+3)]. We have no way of knowing that the (2n+3)/[n*(2n+3)] is also smaller than epsilon, much less use it to find some N which can be applied to find the limit of (n+2)/[n*(2n+3)].

Edit: Oh, is it the following? If f(x) < epsilon for some x > N, then given g(x) < f(x) for all x, then g(x) < epsilon?

Here, g(x) is (n+2)/[n*(2n+3)], and we elect to prove the limit of the easier function (2n+3)/[n*(2n+3)].
Click to expand...

I think your two steps are switched, and that's why it didn't make sense to you.

First, we would show that, for all n > 0, we get (n + 2)/(2n^2 + 3n) <= (2n + 3)/(2n^2 + 3n) = 1/n.

Secondly, we show that for all epsilon > 0, there exists a starting point N such that 1/n < epsilon for n >= N. Hence, 1/n converges to 0 in the limit.

This implies (n + 2)/(2n^2 + 3n) converges to 0. Given an epsilon, the same N for 1/n above also works here.
 
Hey everyone I have two questions I need help with

Factor completely;

1. (x+y)^2 - 16a^2 ...ok I don't believe foiling would do much to help me out in this situation, is is the only thing I can do is extract a -4a? Would the answer be (x+y)^2 -4a(4a) be correct?

2. 2a^3 + 3a^2 - 8a - 12... doing this one via grouping I get a^2(2a+3) -4(2a+3). I believe I then write it as (a^2-4)(2a+3)? But I'm not sure if that's correct.

Any help or confirmation of the answers would be of great help and greatly appreciated
 
Allonym said:
Hey everyone I have two questions I need help with

Factor completely;

1. (x+y)^2 - 16a^2 ...ok I don't believe foiling would do much to help me out in this situation, is is the only thing I can do is extract a -4a? Would the answer be (x+y)^2 -4a(4a) be correct?

2. 2a^3 + 3a^2 - 8a - 12... doing this one via grouping I get a^2(2a+3) -4(2a+3). I believe I then write it as (a^2-4)(2a+3)? But I'm not sure if that's correct.

Any help or confirmation of the answers would be of great help and greatly appreciated
Click to expand...

While you can certainly factor 4a from 16a^2, here, the problem is asking you to use the difference of squares formula, i.e. for any real numbers x and y, we have x^2 - y^2 = (x + y)*(x - y). See how you can apply the formula to the given expression.

Your answer for #2 is correct. You can expand the expression (a^2 - 4)*(2a + 3) to see whether you get back the original polynomial.
 
All the math in this thread looks way too complicated for what I'm working with. Could anybody try answering/explaining this question for me, it was on an exam and I had no idea what I should have been doing to answer it. I assume it has something to do with compound interest/future value because that's what we were working with but I didn't know what formula to apply.
You're looking to buy a car and know you can only afford $300 per month payments. If you get a 5 year loan at 5% interest rate what is the price of the car you can afford?
Click to expand...
 
Tainted Giraffe said:
All the math in this thread looks way too complicated for what I'm working with. Could anybody try answering/explaining this question for me, it was on an exam and I had no idea what I should have been doing to answer it. I assume it has something to do with compound interest/future value because that's what we were working with but I didn't know what formula to apply.
Click to expand...

You're looking to buy a car and know you can only afford $300 per month payments. If you get a 5 year loan at 5% interest rate what is the price of the car you can afford?
Click to expand...

Okay, so a five year loan and you're paying $300 a month, so you're paying a total of 300 * 5 * 12 = 18,000.

The exam question doesn't make it clear if the "5% interest rate" is on the loan principal or per year compounded over the life of the loan.

If it's 5% on the loan, then 18,000/1.05 = 17,142.86 (the rest of the money you paid was interest). So the most expensive car you can buy at that interest rate and that monthly payments is this amount.

If it's 5% per year, then the first step is to get the compounded interest over the life of the loan: 1.05^5 = 1.2763. 18,000/1.2763 = 14,103.28. Anything more expensive than that and your monthly payments would be too high.
 
kgtrep said:
While you can certainly factor 4a from 16a^2, here, the problem is asking you to use the difference of squares formula, i.e. for any real numbers x and y, we have x^2 - y^2 = (x + y)*(x - y). See how you can apply the formula to the given expression.

Your answer for #2 is correct. You can expand the expression (a^2 - 4)*(2a + 3) to see whether you get back the original polynomial.
Click to expand...

Thanks for your help, after racking my brain a little bit I understood what you meant. I ended up getting x^2+2xy+y^2-16a^2, which is the expanded form of the original problem. I believe the final answer is (x+y+4a)(x+y-4a). Hopefully that's correct.
 
Stumpokapow said:
Okay, so a five year loan and you're paying $300 a month, so you're paying a total of 300 * 5 * 12 = 18,000.

The exam question doesn't make it clear if the "5% interest rate" is on the loan principal or per year compounded over the life of the loan.

If it's 5% on the loan, then 18,000/1.05 = 17,142.86 (the rest of the money you paid was interest). So the most expensive car you can buy at that interest rate and that monthly payments is this amount.

If it's 5% per year, then the first step is to get the compounded interest over the life of the loan: 1.05^5 = 1.2763. 18,000/1.2763 = 14,103.28. Anything more expensive than that and your monthly payments would be too high.
Click to expand...
Thanks! I must have fallen a sleep or something because I don't remember covering this in class. :/
 
Allonym said:
Thanks for your help, after racking my brain a little bit I understood what you meant. I ended up getting x^2+2xy+y^2-16a^2, which is the expanded form of the original problem. I believe the final answer is (x+y+4a)(x+y-4a). Hopefully that's correct.
Click to expand...

Yep. And I just realized for #2, you should write (a + 2)(a - 2) instead of (a^2 - 4), as the problem asked us to factor completely, i.e. until we can't any longer using simple terms. Sorry about that.
 
Hmm, I'm somewhat confused with this problem

(3-2i) - (7-4i)...for this problem I realize that I have to distribute the negative but from there do I multiply (3-2i) and (-7+4i) or do I simply combine like terms? On my exam, I multiplied and got it wrong but the instructor never went into why it was wrong, just red ink going through it. So is the answer 13(-1+2i) or is it 2(-2+i)? Any help is appreciated!

kgtrep said:
Yep. And I just realized for #2, you should write (a + 2)(a - 2) instead of (a^2 - 4), as the problem asked us to factor completely, i.e. until we can't any longer using simple terms. Sorry about that.
Click to expand...

No, it's perfectly ok. I appreciate you even taking the time to help me out.
 
Allonym said:
Hmm, I'm somewhat confused with this problem

(3-2i) - (7-4i)...for this problem I realize that I have to distribute the negative but from there do I multiply (3-2i) and (-7+4i) or do I simply combine like terms? On my exam, I multiplied and got it wrong but the instructor never went into why it was wrong, just red ink going through it. So is the answer 13(-1+2i) or is it 2(-2+i)? Any help is appreciated!.
Click to expand...

Just work with the real and imaginary parts seperately. So you have (3 - (-7)) + i((-2) - (-4)) = 10 + 2i so the answer is neither.
 
FortuneFaded said:
Just work with the real and imaginary parts seperately. So you have (3 - (-7)) + i((-2) - (-4)) = 10 + 2i so the answer is neither.
Click to expand...

You sure? I keep getting -4 +2i? I just took 3-2i -7+4i and combined like terms and got -4+2i. I put it into google and it gave me the same answer. If that's correct, can I simplify it and draw out a 2, or I won't be able to because of the imaginary number?
 
Allonym said:
You sure? I keep getting -4 +2i? I just took 3-2i -7+4i and combined like terms and got -4+2i. I put it into google and it gave me the same answer. If that's correct, can I simplify it and draw out a 2, or I won't be able to because of the imaginary number?
Click to expand...

We do get -4 + 2*i by adding the real numbers together and the imaginary numbers together.

You could factor out a 2 and write 2(-2 + i) if you wish. We are simply using the distributive property of numbers (even with the FOIL examples).
 
Allonym said:
You sure? I keep getting -4 +2i? I just took 3-2i -7+4i and combined like terms and got -4+2i. I put it into google and it gave me the same answer. If that's correct, can I simplify it and draw out a 2, or I won't be able to because of the imaginary number?
Click to expand...

Your answer is correct, other poster is wrong.
 
Suppose we have a function f: A -> B. Let C be a subset of A and D be a subset of B.

Then it is not true in general that f(f-inverse(D)) = D and also not true that f-inverse(f(C)) = C?
 
Kieli said:
Suppose we have a function f: A -> B. Let C be a subset of A and D be a subset of B.

Then it is not true in general that f(f-inverse(D)) = D and also not true that f-inverse(f(C)) = C?
Click to expand...
If D is not a subset of f(A), then there's nothing you can apply f to and get D, so the first one isn't true.
If f(a) equals some fixed b for every a in A, then f inverse of any subset of B will be either all of A or the empty set, so the second one isn't true either.
 
Therion said:
If D is not a subset of f(A), then there's nothing you can apply f to and get D, so the first one isn't true.
If f(a) equals some fixed b for every a in A, then f inverse of any subset of B will be either all of A or the empty set, so the second one isn't true either.
Click to expand...

Thanks Therion. I constructed some counter-examples to convince myself (I also drew a simple diagram).

I have a general question.

Suppose we have the following defective matrix [a b] where a and b are column vectors.
a = <3 -4> and b = <1 -1>.

I can find the eigenvalue and eigenvector of this matrix (1 and <-1 2>). However, I also want to find a generalized eigenvector. How do I go about that?
 
Kieli said:
Thanks Therion. I constructed some counter-examples to convince myself (I also drew a simple diagram).

I have a general question.

Suppose we have the following defective matrix [a b] where a and b are column vectors.
a = <3 -4> and b = <1 -1>.

I can find the eigenvalue and eigenvector of this matrix (1 and <-1 2>). However, I also want to find a generalized eigenvector. How do I go about that?
Click to expand...
If A is your original matrix, set B=A-I. Presumably you found the first eigenvector by solving the homogeneous system associated with B. Set v to be this eigenvector.

Then you can find a generalized eigenvector by solving the system Bw=v. Since Bv=0, this will mean that

1) B^2w=Bv=0, and
2) Bw is not itself zero.

These are exactly the properties that you want from a generalized eigenvector.
 
I'm trying to reverse engineer a formula that a certain game I'm playing uses. The way it works is: You have a bunch of gear and other stuff that has a stat called Critical Damage. It's a number with 1 decimal place. The sum of all your Critical Damage stats is converted by some internal formula into a Critical Damage Percentage. The function to do this appears to be roughly linear in terms of your level. I say roughly because I can't apply a straight line function to it. I'm pretty much 100% certain this has to do with rounding error, but I can't figure out where to round / floor in order to make it work out correctly.

I've observed the following set of Crit Damage / Crit Dmg% pairs at various levels:

Code: 
Level    Critical Damage     CRIT DMG%
  76           50              39.70%
  75           50              40.00%
  74           50              40.30%
  73           50              40.70%
  72           50              41%
  71           50              41.30%
  70           50              41.70%
  69           50              42.00%
  68           50              42.40%
  67           50              42.70%
  66           50              43.10%
  65           50              43.50%
  64           50              43.90%
  63           50              44.20%
  62           50              44.60%
  61           50              45.00%
  60           50              45.50%
  59           50              45.90%
  58           50              46.30%
  57           50              46.70%
  56           50              47.20%
  55           50              47.60%
  54           50              48.10%
  53           50              48.50%
  52           50              49.00%
  51           50              49.50%
  50           50              50.00%

From level 50 and below the CRIT DMG% is always exactly 50%, so it's not important.

The problem is that I don't know where the floors go in the formula, and it's already been truncated in the observed CRIT DMG% values so solving for the slope gives an incorrect value.

The ultimate goal here is to get f(Level, CriticalDamage) = CritDmgPct
 
cpp_is_king said:
Code: 
Level    Critical Damage     CRIT DMG%
  76           50              39.70%
  75           50              40.00%
  74           50              40.30%
  73           50              40.70%
  72           50              41%
  71           50              41.30%
  70           50              41.70%
  69           50              42.00%
  68           50              42.40%
  67           50              42.70%
  66           50              43.10%
  65           50              43.50%
  64           50              43.90%
  63           50              44.20%
  62           50              44.60%
  61           50              45.00%
  60           50              45.50%
  59           50              45.90%
  58           50              46.30%
  57           50              46.70%
  56           50              47.20%
  55           50              47.60%
  54           50              48.10%
  53           50              48.50%
  52           50              49.00%
  51           50              49.50%
  50           50              50.00%
Click to expand...


I think the person added a sinusoidal function with a low amplitude to a linear function (maybe floor/round afterwards).

Below, I show the plot of critical damage percentage against the level (blue line), and the plot of the linear function y(x) = -0.4x + 69.7 (red line).


We see that the magnitude in the difference between your given data (blue line) and the linear function (red function) is odd-symmetric about level = 55.5 and level = 70.5.

(The slope between the levels 55 and 56 and between the levels 70 and 71 is -0.4, which gives us the y-intercept of 69.7.)


We can imagine a sinusoidal function whose minimum occurs around level = 62. I'm not sure why it didn't at 63 instead, but eh...


dEFs1bT.png
Click to expand...
 
kgtrep said:
I think the person added a sinusoidal function with a low amplitude to a linear function (maybe floor/round afterwards).

Below, I show the plot of critical damage percentage against the level (blue line), and the plot of the linear function y(x) = -0.4x + 69.7 (red line).


We see that the magnitude in the difference between your given data (blue line) and the linear function (red function) is odd-symmetric about level = 55.5 and level = 70.5.

(The slope between the levels 55 and 56 and between the levels 70 and 71 is -0.4, which gives us the y-intercept of 69.7.)


We can imagine a sinusoidal function whose minimum occurs around level = 62. I'm not sure why it didn't at 63 instead, but eh...
Click to expand...

Imagine you had something of the form:

y = &#8970;1000*(mx + b)&#8971;/1000

The factor of 1000 here is essentially means "truncate to 3 decimal places"

Wouldn't this exhibit that same sinusoidal behavior? For example, I just plotted this up:


cbAsFGC.png



So I actually think the formula is something like the above perhaps. But the problem is how to solve for m and b when this floor is in the way? I know that in general for &#8970;x&#8971; = y means y <= x < y+1. So I tried applying that here.

1000y = &#8970;1000*(mx + b)&#8971;
1000y <= 1000*(mx + b) < 1000y+1
y <= mx + b < y + .001

But now what do I do? Normally you would solve for slope by doing (y2-y1)/(x2-x1). Does the same hold true here, where I just subtract 2 entire inequalities from each other to come up with bounds on m?

For example, using 76 and 50, I would have two equations:

.397 <= 76m + b < .398
.5 <= 50m+b < .501
 
Maybe I'm over complicating this. Perhaps I just need to do a least squares curve fit on the observed data and round the result to 1 decimal place.

Edit: Wow, yea that works great. I used a nonlinear least squares fit with ax^2 + bx + c, and it came up with: 0.003528x^2 - 0.8388x + 83.087. Rounding this to one decimal place matches every value. This should be good enough.
 
cpp_is_king said:
Maybe I'm over complicating this. Perhaps I just need to do a least squares curve fit on the observed data and round the result to 1 decimal place.

Edit: Wow, yea that works great. I used a nonlinear least squares fit with ax^2 + bx + c, and it came up with: 0.003528x^2 - 0.8388x + 83.087. Rounding this to one decimal place matches every value. This should be good enough.
Click to expand...

Yeah, I checked that works. Neat. My linear + sinusoidal model could only match 22 points. :(
 
kgtrep said:
Yeah, I checked that works. Neat. My linear + sinusoidal model could only match 22 points. :(
Click to expand...

Gah, while this solves the question i asked, it doesn't solve my original problem. Basically, the curve fit produces a formula that is designed to solve the case of x=50. It is still a little bit off for other values.

What I thought would work would be to compute the value at x=50, then for some value of x other than 50, just do x * f(50) / 50.

But I'm still off between .1% and .2% from time to time.

Boo. Maybe I need to do a curve fit with multiple independent variables, but I will probably need quite a large sample size with multiple values at each level for that.
 
Hi MathGAF!

How do I derive:
Code: 
(3x^2)/(1+x^3)

Wolfram gives me:
Code: 
-(3 x (-2+x^3))/(1+x^3)^2

But I could use a decent step-by-step. :(
 
CoolOff said:
Hi MathGAF!

How do I derive:
Code: 
(3x^2)/(1+x^3)

Wolfram gives me:
Code: 
-(3 x (-2+x^3))/(1+x^3)^2

But I could use a decent step-by-step. :(
Click to expand...

This is the quotient rule. If f = u(x)/v(x) then f' = (u'v - uv')/v^2. (' implies differentiation by x)

So u = 3x^2, v = (1+x^3), u' = 6x, v' = 3x^2

So f = [(6x)(1+x^3) - (3x^2)(3x^2)]/(1+x^3)^2 after substitution.

= -(3 x (-2+x^3))/(1+x^3)^2 after simplifying the numerator.
 
FortuneFaded said:
This is the quotient rule. If f = u(x)/v(x) then f' = (u'v - uv')/v^2. (' implies differentiation by x)

So u = 3x^2, v = (1+x^3), u' = 6x, v' = 3x^2

So f = [(6x)(1+x^3) - (3x^2)(3x^2)]/(1+x^3)^2 after substitution.

= -(3 x (-2+x^3))/(1+x^3)^2 after simplifying the numerator.
Click to expand...

Cheers for the quick answer! I had screwed up the formula in my head. :)

Anyway, I'll try one last quick question. I have a function where I'm finding and classifying stationary points.

http://www.wolframalpha.com/input/?i=f(x)=x^6-5x^4

Now, the second derivative becomes 30x^4-60x^2, which if I put in one stationary point I got from f'(x), = 0, gives me f''(x) = 0, but that would indicate a terrace point, despite it being a maximum? Is it just that it's in 0 that's screwing me?
 
CoolOff said:
Cheers for the quick answer! I had screwed up the formula in my head. :)

Anyway, I'll try one last quick question. I have a function where I'm finding and classifying stationary points.

http://www.wolframalpha.com/input/?i=f(x)=x^6-5x^4

Now, the second derivative becomes 30x^4-60x^2, which if I put in one stationary point I got from f'(x), = 0, gives me f''(x) = 0, but that would indicate a terrace point, despite it being a maximum? Is it just that it's in 0 that's screwing me?
Click to expand...

If I were to classify the point, I would call it a maximum point (you can prove that by inserting an arbitrarily small positive or negative number).
 
Majestad said:
Could you guys lend a brother a hand:

eBePUWx.png


I'm stuck.
Click to expand...

The two parts of f(x) for x < -7 and x>= -7 are both continuous (since both are polynomials). The critical point is at x = -7. In order for f(x) to be continuous there, it must satisfy


Code: 
   lim             
x -->  -7,      f(x)   =    f(-7)
x  <  -7

or in other words

Code: 
   lim             
x -->  -7,      mx - 12  =    30.
x  <  -7

Now determine m such that the above is satisfied (I hope you are familiar with lim notation).
 
Majestad said:
Could you guys lend a brother a hand:

eBePUWx.png


I'm stuck.
Click to expand...

You know m*x - 12 goes from almost -7 to negative infinity.

You also know that x^2 + 2x - 5 goes from -7 to infinity. So you just need to connect almost negative seven to negative seven.

While it seems like you have two unknowns, you actually only have one, because you can put x=-7 into both equations as that is the only point it could possible be discontinuous.
Answer below. Try it before you unspoil it : )

Thus m*-7 -12 = -7^2 + 2*-7 -5
Using simple arithmatic:
m*-7 - 12 = 49 - 14 -5
m*-7 = 49 - 14 - 5 + 12
m*-7 = 42
m = 42/-7
And thus m = -6

You can test this by plugging both equations
-6x-12
and x^2 + 2x -5 and confirming they do indeed intersect at x = -7.

Is this for Calculus or pre-calculus? If it's for Calculus you could (and should) definitely do it with limit notation above. If you've never heard of a limit before, this works as well.

Edit: Oh shit I'm blind. Thanks KP. Fixed.
 
Xe4 said:
You know m*x - 12 goes from almost 7 to negative infinity.

You also know that x^2 + 2x - 5 goes from 7 to infinity. So you just need to connect almost seven to seven.

While it seems like you have two unknowns, you actually only have one, because you can put x=7 into both equations as that is the only point it could possible be discontinuous.
Answer below. Try it before you unspoil it : )

Thus m*7 -12 = 7^2 + 2*7 -5
Using simple arithmatic:
m*7 - 12 = 49 + 14 -5
m*7 = 49 + 14 - 5 + 12
m*7 = 70
m = 70/7
And thus m = 10

You can test this by plugging both equations
10x-12
and x^2 + 2x -5 and confirming they do indeed intersect at x = 7.
Click to expand...

While your explanation is correct, the critical point is at x = -7, not at x = 7. So your calculation is wrong.
 
Karma Police and Xe4, thanks a lot. You guys are rockstars. Yeah this is my first week of Calculus so is a bunch of stuff I haven't seen before. I'll be studying over the weekend to get more comfortable with it since right now it feels kinda scary lol.
 
Given the Airy equation y'' - xy = 0, find a series solution centered at x = 1.

I need help with identifying ordinary points. I know the definition of an ordinary point is one such that p(x) and q(x) are analytic at these points.

But that doesn't help me find them.

If I can identify the ordinary points, then solving the diffyEQ is rather elementary.

Thanks for any help!
 
Kieli said:
Given the Airy equation y'' - xy = 0, find a series solution centered at x = 1.

I need help with identifying ordinary points. I know the definition of an ordinary point is one such that p(x) and q(x) are analytic at these points.

But that doesn't help me find them.

If I can identify the ordinary points, then solving the diffyEQ is rather elementary.

Thanks for any help!
Click to expand...

Hmm, I'm not sure if I help but isn't x=1 your ordinary point and don't you just need to change your variable to t=x-1? Then y(x)=y(t+1)=u(t) and you get u''+(t+1)u=0 for t=0, which should be easy.
 
I fucked up this question in a quiz because I ran out of time

given f(x) = blah and g(x) = blah, find f(g(y))

was I supposed to solve for x in g(y) and f(g(y))?
 
ohlawd said:
I fucked up this question in a quiz because I ran out of time

given f(x) = blah and g(x) = blah, find f(g(y))

was I supposed to solve for x in g(y) and f(g(y))?
Click to expand...

No, you were supposed to replace x in f(x) with the entire equation for g(x). Then simplify.

For example, f(x) = 3x^2 + 7, g(x) = (x+2)^2 + 1

f(g(y)) = 3[(y+2)^2 + 1]^2 + 7
= 3 y^4+24 y^3+78 y^2+120 y+82
 
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