Objectivism Online Forum

Rate this topic

## Recommended Posts

I am having problems with a math assignment. The class is 100 level.

When f(x) = -6x+5 and g(x) = (3x^2)-10 I am supposed to find (fg)(x), (g+f)(x), and 3*g(x-4)

I am then supposed to simplify the results.

For (fg)(x) I came up with (-6x+5)((3x^2)-10)

For (g+f)(x) I came up with (-6x+5)+((3x^2)-10)

For 3*g(x-4) I came up with 3((3(x-4)^2)-10)

So far so good (I think), but I can't remember how to simplify any of the results. Math is not my interest and sometimes it frustrates me.

Can anyone help?

##### Share on other sites

While I'm at it, can anyone explain domains and ranges of functions to me? The concept confuses me. ##### Share on other sites

I got (18x^3)+60x+(15x^2)-50 for the first problem. The original equation contained a negative 6x rather than a positive 6x.

I got (3x^2)+6x-5 for the second problem. Did I make an error?

For the third problem I arrived at the same answer.

Anyway, thank you very much for the help.

##### Share on other sites
Normally I charge for this; but since I don't have a way for you to pay me online, I'll just take pity on you an tell you the answer for free. [...]

Thanks for the free tutotial. ##### Share on other sites

Are you sure the first problem was (fg)(x) and not f(g) of (x)? Because they are two different things and I have never encountered a problem asking for (fg)(x). But that could just be me...

Anyway, I also have a math question.

Does anyone know how to find the area of two circles whose centers are one unit apart and have both have a radius of 1?

It is on my problem set, and it has me really stumped. Drawing traingles doesn't work, the arc measurements don't help, the angle measurements don't help. ARRRRGH! I spent about an hour just staring at it and thinking. No dice. Any help would be sincerely appreciated.

edited to fix a large amount of typos ##### Share on other sites
Does anyone know how to find the area of two circles whose centers are one unit apart and have both have a radius of 1?

edited to fix a large amount of typos OK, the two circles are overlapping externally and their centers are on each other(since both centers are 1 apart and have radius 1. Let the two centers be O and O'.

Let the two circles intersect at A(lower point) and B(upper point). Now to find area of both circles together find area of two circles and subtract the area of the two secants in between.

To find the area of the two secants, join AB, AO, BO, AO', BO' and OO' (meeting AB at D). Triangles AOB and AO'B are congruent(same radii and AB=AB). Therefore triangles OBD and O'BD are also congruent(SAS congruence)

OD=O'D=1/2*OO'=.5

In AOD, OD is .5 and OA is 1. Applying trig, angle AOD is 60 and angle AOB is 120. Find area of sector with angle 120 and subtract the area of triangle AOB. Multiply the area by two and subtract it from the area of the two circles(2*pi) taken separately. You get the area.

Hope this helps Edited to fix typos

Edited by tommyedison

##### Share on other sites

OMG thank you so much. Geometry was my worst math class, I much prefer algebra. I knew it had to have something to do with the angles but... I just had no idea. Thanks again. EDIT: My fellow Mathematical Investigations Three students say thank you too. BTW- Your profile says you were born in 1990. What grade are you in? Just curious... I'm a sophomore. ##### Share on other sites
I got (3x^2)+6x-5 for the second problem. Did I make an error?

Anyway, thank you very much for the help.

No the origional problem contained a -10 and +5 which equal -5. Necessary Truth had a +10 and +5. No harm done.

##### Share on other sites
OMG thank you so much. Geometry was my worst math class, I much prefer algebra.  I knew it had to have something to do with the angles but... I just had no idea. Thanks again. EDIT: My fellow Mathematical Investigations Three students say thank you too. BTW- Your profile says you were born in 1990. What grade are you in? Just curious... I'm a sophomore. You're welcome. Thanks for the great math problem.

I'm a sophomore too. Edited by tommyedison

##### Share on other sites
You're welcome. Thanks for the great math problem.

I'm a sophomore too. Wow. What kind of school lets you take BC Calculus before high school? (You said you took it two years ago, right?) Is it just a regular public school or a private one? The highest my school (public) let us go is Algebra in seventh grade. Well, actually one of my friend's mom talked to the school and he took it in sixth grade, but I didn't know that until too late. What math are you in now?

-Zak

##### Share on other sites
Wow. What kind of school lets you take BC Calculus before high school? (You said you took it two years ago, right?) Is it just a regular public school or a private one? The highest my school (public) let us go is Algebra in seventh grade. Well, actually one of my friend's mom talked to the school and he took it in sixth grade, but I didn't know that until too late. What math are you in now?

I am in a private school in India. My school doesn't give classes in BC, I just prepared on my own and took the exam.

Now I doing the math classes at my school, things which I have already done biding my time waiting for college.

P.S. I took the BC last year.

Edited by tommyedison

##### Share on other sites

It was the same at my old school. We weren't allowed to take Algebra until eigth grade. Freshman year was geometry. But I transferred to a pseudo public school for geeks so now we can go all the way to Differential Equations, Multivariable Calculus etc. So I get to start BC calc next year. This year I'm in what they call Mathematical Investigations. It's sort of like Algebra 2 and Trig. combined. I think I'll take Dif. EQ my senior year. I really love math, and I'm supposed to be good at it, but wow, BC calc as an 8th grader? All I can say is wow... That is totally awesome.

EDIT: As a ninth grader is still totally awesome. ##### Share on other sites

Yeah, that is pretty frickin' amazing. I'm in Pre-calc now, but its kinda boring. Like I get the stuff, but make really stupid mistakes so its really frustrating; plus last year (Alg II/Trig.) I had the best teacher! He is the real calculus teacher and a serious genius. We speak the same language, so much so that I didn't even have to listen after the first five minutes. It just all kinda clicked.

Non-contradictor: How can you start BC before you take AB? My school makes us take them in order. AB then BC.

Zak

##### Share on other sites
It was the same at my old school. We weren't allowed to take Algebra until eigth grade. Freshman year was geometry. But I transferred to a pseudo public school for geeks so now we can go all the way to Differential Equations, Multivariable Calculus etc. So I get to start BC calc next year. This year I'm in what they call Mathematical Investigations. It's sort of like Algebra 2 and Trig. combined. I think I'll take Dif. EQ my senior year. I really love math, and I'm supposed to be good at it, but wow, BC calc as an 8th grader? All I can say is wow... That is totally awesome.

How can you take Differential Equations in your senior year and BC next year. Dif. EQ are a part of Calculus.

EDIT: As a ninth grader is still totally awesome. Yeah, that is pretty frickin' amazing.

Thanks ##### Share on other sites
How can you take Differential Equations in your senior year and BC next year. Dif. EQ are a part of Calculus.

Thanks You know it is sad by the time you were a freshman in high school you had gotten further along then I ever did. Alg II. ##### Share on other sites
Non-contradictor: How can you start BC before you take AB? My school makes us take them in order. AB then BC.

Zak

The best answer I have is that my school does not follow the normal patterns. The class I am in now is sort of part calculus part trig and part algebra 2. So we have enough understanding to jump straight into BC. Hardly anyone here ever takes AB, since according to my counselor, they are pretty much the same class, BC is just harder. I guess it's kind of like Pre-Algebra and Algebra. One goes into more depth. My school is very weird...

##### Share on other sites
How can you take Differential Equations in your senior year and BC next year. Dif. EQ are a part of Calculus.

I dunno, I didn't set up the classes. Dif. EQ is an elective, so my guess is that the Dif EQ class goes into *much* greater depth on the subject than BC calc. But I don't really know. They've got to talk about something all semester though...

##### Share on other sites

Okay, I have a math question of my own now. I need to prove that 1+tan*x=sec*x. (*=squared. It's just easier)

Can I use the other pythagorean identities to do this, as long as I don't use this one?

If I do:

1+sin*x/cos*x = 1/cos*x

1=(1/cos*x)-(sin*x/cos*x)

1=(1-sin*x)/cos*x

1=cos*x/cos*x

1=1

Right?

##### Share on other sites

Thank you. I actually just took the test and, even though he told us explicitly we would have to do a proof of one of the Pythagorean Identities, it wasn't on the test. The test actually turned out to be the easiest one all year. Thanks for the help though, who knows when I will have to do it again.

I never would have thought of how to do that. Zak

##### Share on other sites

I really like this thread. I have another math problem that has me stumped:

"A rectangle has two vertices on the x-axis and two vertices above the x-axis on the curve y=36-x^2.

a) write a function describing the area of the rectangle, A(x)

b)graph the function

c) what is the maximum possible area?"

Hmmm... Any help would be greatly appreciated. edit to remove an extra word-"am"- from the first sentence

##### Share on other sites

I hope I'm not doing your hw here...

a) The function is given by 2*abs(x)*f(x) or for graphing 2*sqrt(x^2)*f(x) (base times height)

b)graph it

c) pull the max area off the graph.

QED

Tell me why the answer to a) is what it is.

Edit: where f(x) = 36-x^2

Also, if you know how to differentiate, you can find the extrema of A(x) easily.

##### Share on other sites
I hope I'm not doing your hw here...

a) The function is given by 2*abs(x)*f(x) or for graphing 2*sqrt(x^2)*f(x) (base times height)

b)graph it

c) pull the max area off the graph.

QED

Tell me why the answer to a) is what it is.

Edit: where f(x) = 36-x^2

Also, if you know how to differentiate, you can find the extrema of A(x) easily.

You're not doing my homework, If I turned in just an answer, I'd get no credit. I was asking how to do the problem. It is a problem off my weekly problem set.

If I knew why that is the answer, I probably wouldn't have needed to ask in the first place. Why do you multiply it by x? Is that the base or the height? I am so confused . I really wasn't asking for the answer, just how to do it.

I do not know how to "differntiate" (or at least I don't think I do).

Thank you for your help. ##### Share on other sites

Okay dude, I'll give it a whirl.

The equation 2abs(x) * (36-X^2) is the equation of the base times the height (or length time width). You go x coordinates laterally. You multiply that by 2 because its only half of the total base lenth. (Its absolute value because you can't have a negative side length) You multiply by f(x) (because 36-x^2 gives you the y-cordinate for a specific x value) to get the total area.

Make sense?

Zak

Edited by realitycheck44

##### Share on other sites

Oh, okay, in that case, let me explain. (By the way, I was joking about the hw, I don't expect dishonesty on this board!)

Do you know what the graph of 36-x^2 looks like? It's an upside down "U" with the peak at y = 36. The zeros, or roots, of y=36-x^2 are +6 and -6. This is a symmetric, even function.

Now, picture your two veritices on the horizontal axis, one at +x and the other at -x (symmetric to make it easy), and then picture two vertical lines emanating from these vertices. They will, above the x-axis, intersect your upside down U at two points. Where do they intersect? Precisely at f(x) or f(-x), because your function is an even function (f(x) = f(-x)).

Now, how do you find the area of a rectangle? Base*height. So, in this case, your base is given by the distance between your two vertices, -x and +x, which is 2*x. What is the height? It is given by the point where those vertical lines emanating from x and -x intercept your function f(x), but this is precisely f(x) (or f(-x)! So base*height = (2*abs(x))*(f(x)), where abs(x) means the absolute value of x. I use this because x can take negative values, and you don't want a negative area!

Let me know if you have further questions. I've included a graphic for your viewing pleasure!

<center><img src="http://www.d-anconia.com/images/plot.gif"></center>

Those two vertical lines emanate from +x and -x, and touch your function at f(x) and f(-x), which are equal.

##### Share on other sites

Okay, I think I get it now. Thank you guys for your help. I was having trouble visualizing how to get the base, f(x) is the height, because the height is y if it ends on the x-axis. Then the base is two times x because x is only half of the base. x has to be only half of the base because f(x) is symmetrical over the y-axis and so you can only get a rectangle if the two points are an equal distance from the y-axis.

Is this right?

Thanks again for your help. ## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account. ×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.