/sci/ - Science & Math

[Anonymous No.10591216]

Let's see how good /sci/ actually is at math.

[Anonymous No.10591221]

>>10591216
>let
>let
>let

[Anonymous No.10591225]

>>10591216
>>10591216
if g(x) = f(x) then a=b.
that means h(x)=2f(x)

therefore i think OP fucked up because the statement is only true if you substitute “+” for “-“ in the last line

[Anonymous No.10591226]

what the fuck, it doesn't though

[Anonymous No.10591233]

>>10591225
Or x is zero and a = -b

It’s clear, then, that this is the minimum y value, and so the slope must be zero

[Anonymous No.10591235]

>>10591225
>g(x) = f(x) then a=b
i think op meant it as in, where the the functions meet, not when they are literally the same thing

[Anonymous No.10591236]

op here, just realized how retarded i am.
I need to re learn the actual terminology.

[Anonymous No.10591258]

>>10591216
>>10591236
It's pretty close. I can tell what you meant, assuming the 3 lets are supposed to be for all x in R and the last line is just if g(x) = f(x) then h'(x) = 0.

[Anonymous No.10591279]

>>10591258
does my proof at least make sense though

[Anonymous No.10591280]

But that's fucking wrong.

[Anonymous No.10591285]

>>10591216
Rolle's lemma.

[Anonymous No.10591290]

>>10591216
>>10591279
No, because it's false. Let a = 0 and b = 0, then [math]f = g = x \mapsto x^2[/math],[math]h = x \mapsto 2x^2[/math], and [math]f(1) = g(1)[/math]. But [math]h'(1) = 4 \neq 0[/math]

[Anonymous No.10591297]

>>10591279
You divided by a - b, but that could be 0.

[Anonymous No.10591311]

>>10591290
I don't understand. This is probably me misusing terminology though.

When i plot it, the turning point of [math]h(x)[/math] is always on the same [math]x[/math] coordinate as the intersection of [math]f(x)[/math] and [math]g(x)[/math].

Would that not make [math]h'(x) = 0[/math] wherever [math]f(x) = g(x)[/math]?

[Anonymous No.10591317]

>>10591311
Did you read the post you replied to?

[Anonymous No.10591327]

>>10591279
The proof is okay for x not equal to zero and a-b =/= 0, but those cases can be easily proven on their own

[Anonymous No.10591329]

>>10591327
The case that a-b=0 cannot be proven

[Anonymous No.10591334]

>>10591317
I did, and when what i said still applies.

Regardless of what values, i plug in the turning point of h(x) is always on the same x coordinate as the intersection of f(x) and g(x).

[Anonymous No.10591338]

Yeah OP meant that a != b.

This is obviously true since f(x) and g(x) are just horizontal translations of x^2. At their point of intersection the slopes of both parabolas are equal in magnitude but with opposite signs, d/dx f(x) = -d/dx g(x) so d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x) = 0.

[Anonymous No.10591339]

>>10591334
>turning point of h(x) is always on the same x coordinate as the intersection of f(x) and g(x)
No, the intersection might contain more than 1 point, but the turning point is always 1 point.

[Anonymous No.10591343]

>>10591216
>Let's see how good /sci/ actually is at math.
Underage retard posting his homework

[Anonymous No.10591347]

>>10591339
You're right, so the proof is only valid for when a - b != 0, but outside that it should work, or am i still missing something?

[Anonymous No.10591363]

>>10591216
OP is the one bad at math
POTTERY

[Anonymous No.10591366]

>>10591363
pls forgive me

[Anonymous No.10591371]

>>10591366
please don't ask anonymously

[Anonymous No.10591380]

>>10591366
You're not OP.

[Anonymous No.10591389]

>>10591329
Are you braindead?
The case a - b = 0 implies that g(x) = f(x) occurs only when a = b = 0. Therefore h(x) = 2x^2, and we get h'(x) = 0.

[Anonymous No.10591398]

>>10591389
Wow, you're even dumber than OP

[Anonymous No.10591407]

>>10591389
>Therefore h(x) = 2x^2, and we get h'(x) = 0.
>Therefore h(x) = 2x^2, and we get h'(x) = 0.

[Anonymous No.10591416]

>>10591407
I think he's trying to say a - b = 0 implies that g(x) = f(x) occurs only when x = 0

[Anonymous No.10591434]

Do you morons even understand that h'(x) is never going to be 0 for x not equal to 0

[Anonymous No.10591437]

>>10591416

yeah.. I'm going to have to say, if op's problem had stated that a cannot equal b, then the only way f(x) could equal g(x) would be if x equaled zero, in which case h'(x) would equal zero.

Otherwise I can't make this work. There is no other way to make f(x) + g(x) equal a constant so that h'(x) is zero...

Just sayin'. The problem should have stated that a cannot equal b.....

[Anonymous No.10591445]

>>10591437
>the only way f(x) could equal g(x) would be if x equaled zero
No retard. Is this >>10591389 you?
>There is no other way to make f(x) + g(x) equal a constant so that h'(x) is zero...
I think you don't understand the problem. It's not saying that h' is the constant 0 function

[Anonymous No.10591446]

>>10591434
>h'(x) is never going to be 0 for x not equal to 0
Wrong

[Anonymous No.10591462]

oooh I see what OP was trying to do

So let's say f(x) = (x+2)^2 and g(x) = (-x-2)^2

when you take the derivative and add them you will get zero.
of course it doesn't work because the negative gets u-subbed in the g(x) derivative

[Anonymous No.10591463]

>>10591446

Wrong

[Anonymous No.10591467]

>>10591462
>g(x) = (-x-2)^2
Where do you see -x, retard?

[Anonymous No.10591477]

>>10591416
That's wrong. a - b = 0 implies a = b implies f(x) = g(x) for all x.

[Anonymous No.10591478]

>>10591463
Choose a = b = 1. Now where is h' 0?

[Anonymous No.10591481]

/sci/ - retarded highschoolers

[Anonymous No.10591485]

>>10591477
I know. I was trying to interpret that he was saying in >>10591389, which makes even less sense

[Anonymous No.10591488]

>>10591478

> 2x^2 is 0

are you retarded?

[Anonymous No.10591490]

>>10591488
Who are your quoting, retard?

[Anonymous No.10591584]

I did this in my head and it took 30 seconds to write down. Did this thread really deserve 43 replies?

[Anonymous No.10591605]

>>10591584
Your didn't even understand the question, and you are confusing functions evaluated at x with the function itself

[Anonymous No.10591640]

>>10591584
Another victim, the thread that keeps on giving

[Anonymous No.10591646]

>>10591225
>if g(x) = f(x) then a=b.

[Anonymous No.10591654]

>>10591640
>>10591646
Point out any flaw, mathlets. This is first year calculus

[Anonymous No.10591667]

>>10591654
g(x) = f(x) is not the same as g = f, dumb highschooler

[Anonymous No.10591673]

>>10591654
You have no right to call anyone a mathlet, dumbshit nigger

[Anonymous No.10591690]

why is everybody changing x? you are not supposed to set x to anything just generally show that h'(x)=0 which is never the case except for x=0

is everybody a moron in this thread?

[Anonymous No.10591693]

>>10591667
>>10591673
The functions are only equal when a=b or when b=-2x-a which yields the same derivative. Any other issues that you can't prove yourself?

[Anonymous No.10591707]

>>10591690
It's not for all x, it's if x is such that f(x) = g(x) then h'(x) = 0. You're the moron.
>>10591693
>The functions are only equal when a=b or when b=-2x-a
b is a constant, retard. And yes, the functions are equal only when a=b. But f(x) is not a function and neither is g(x)

[Anonymous No.10591737]

>>10591690
>>10591693
Here, I will state the problem precisely since you guys are having such a hard time grasping it:
Let [math]a,b \in \mathbb R[/math] such that [math]a \neq b[/math], and [math]f,g,h[/math] be real functions given by [math]f=x\mapsto (x+a)^2,\ g = x\mapsto (x+b)^2,\ h = x \mapsto f(x) + g(x)[/math]. Show that for all [math]x\in\mathbb R,\ g(x) = f(x) \implies g'(x) = 0[/math].

[Anonymous No.10591741]

>>10591690
>>10591693
Here, I will state the problem correctly since you guys are having such a hard time grasping it:
Let [math]a,b \in \mathbb R[/math] such that [math]a \neq b[/math], and [math]f,g,h[/math] be real functions given by [math]f=x\mapsto (x+a)^2,\ g = x\mapsto (x+b)^2,\ h = x \mapsto f(x) + g(x)[/math]. Show that for all [math]x\in\mathbb R,\ g(x) = f(x) \implies h'(x) = 0[/math].

[Anonymous No.10591744]

>>10591338
This is an awesome explanation, thanks!

[Anonymous No.10591765]

>>10591707
>b is a constant, retard
Of course b is a constant. I did it that way to demonstrate that the case where x+a=-x-b has no effect on the derivative. That x is the particular x at which the functions intersect though, so maybe that approach was fallacious

[Anonymous No.10591769]

>>10591765
Who are you?

[Anonymous No.10591771]

>>10591741
Okay

>They intersect at a single point (x=-0.5(a+b))
>h(-0.5(a+b)) is a constant
>The derivative of any constant is 0

Done. This is stupider than the problem as originally stated which was wrong

[Anonymous No.10591777]

>>10591771
That is the original problem (minus the a != b part). The problem was solved 8 posts in, and the rest of the posts are just people not understanding the problem and confusing g(x) = f(x) as g = f

[Anonymous No.10591784]

>>10591771
And you still didn't solve it.
>h(-0.5(a+b)) is a constant
Yes, h evaluated at any point is a constant. That doesn't tell you anything above the derivative at that point.

[Anonymous No.10591790]

>>10591777
When a=b, f=g so it's an entirely different problem when you add a!=b

[Anonymous No.10591792]

>>10591771
>>10591784
Pretty pathetic desu, trying to show off your calculus 1 skills only to not know what a derivative is

[Anonymous No.10591794]

>>10591784
Are you high? a and b are constants

[Anonymous No.10591799]

>>10591790
It's ok, you didn't interpret the original problem as what I stated minus the a != b part, you said >>10591646

[Anonymous No.10591804]

>>10591794
Can you not read? Yes, h(-0.5(a+b)) is a constant, so is h(0), h(1), h(2), etc. That doesn't mean the derivative is 0 at those points.

[Anonymous No.10591811]

>>10591771
>>10591794
You are literally just saying [math]h(-0.5(a+b))\in\mathbb R[/math] and therefore [math]h'(-0.5(a+b)) = 0[/math] . But [math]h(x) \in \mathbb R[/math] for all [math]x\in\mathbb R[/math] since it is a real function.

[Anonymous No.10591816]

>>10591799
No, I said f(x)=g(x) when a=b which is true and valid unless you stipulate that a!=b

[Anonymous No.10591828]

>>10591804
>>10591811
>h(x)=-0.5(a+b) for all x
I just wrote it wrong because I pounded it out really quick without thinking

[Anonymous No.10591835]

>>10591816
>No, I said f(x)=g(x) when a=b
Oh, then in that case your proof is just wrong, even for just trying to disprove the misinterpreted problem. You deduced that a=b from "f(x)=g(x) when a=b" and used the fact a=b in the next step of your disproof. But you cannot assume a=b since you have not deduced that. I'm assuming this >>10591584 is you.

[Anonymous No.10591836]

>>10591828
>h(x)=-0.5(a+b) for all x
Fuck me. I typed it wrong again. This is why I didn't want to type it all out.

>h(x)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2

[Anonymous No.10591839]

>>10591828
Yeah, that's wrong too buddy. h is in fact not a constant function, it actually only has a derivative of 0 at 1 point.

[Anonymous No.10591848]

>>10591836
>h(x)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2
That still doesn't say anything about h'(x)

[Anonymous No.10591850]

>>10591835
By inspection there are only two possible cases for g(x)=f(x): a=b and a single point intersection. You can't say that I'm wrong, only that I'm incomplete

[Anonymous No.10591855]

>>10591848
It says that it's 0. Are you high? The derivative with respect to x of any expression that involves only constants is 0

[Anonymous No.10591862]

>>10591279
This looks right you could have cancelled out a (a-b) out earlier on and made it much simpler

[Anonymous No.10591864]

>>10591855
x is a fucking real number, not a bound variable. If it was h(x)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2 for all x in R, then you would be right. How can you so dense, I've already told you this.

[Anonymous No.10591868]

>>10591855
Here, I will prove that the identity function id has a derivative of 0 at 0 using your logic: id(0) = 0 is a constant, therefore id'(0) = 0.

[Anonymous No.10591878]

>>10591864
>h(x)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2
>h(0)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2
>h(-25)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2
>h(30)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2
>h(x)=h(0)=h(-25)=h(30)
Looks to me like h(x) is a constant value for all x and therefore has a derivative of 0

[Anonymous No.10591882]

>>10591868
I think you need to work on your reading comprehension

[Anonymous No.10591890]

>>10591878
h is actually not a constant function. How are you getting that h(0)=h(-25)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2? Did you forget that you got h(x)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2 by plugging x=-0.5(a+b) into the formula for h? Why don't you try graphing h and see if it's a constant function?

[Anonymous No.10591897]

>>10591882
Here: let x = 0, then id(x) = 0. Therefore, id'(x) = 0.

[Anonymous No.10591906]

>>10591890
h is a constant function with the constraints that f(x)=g(x) and a!=b

[Anonymous No.10591912]

>>10591897
That's closer. Have a (you) for your efforts

[Anonymous No.10591914]

>>10591906
>f(x)=g(x)
Do you mean f(x)=g(x) for some x or for all x? It doesn't matter, because h is not constant either way.

[Anonymous No.10591919]

>>10591906
>>10591912
How old are you? Just curious.

[Anonymous No.10591922]

>>10591914
See
>>10591741

You also could have used the backlinks

[Anonymous No.10591929]

>>10591922
I think you don't understand what that means. There is no condition that f(x) = g(x) for all x in R.

[Anonymous No.10591935]

>>10591929
I'm not the one who typed it out, and I didn't look at it too closely the first time around. I think he added "for all x" accidentally

[Anonymous No.10591938]

>>10591922
Btw, have you tried graphing h yet?

[Anonymous No.10591952]

>>10591938
I graphed h long ago. When a=b, h is a non-constant line when a!=b, h is a constant line

[Anonymous No.10591954]

>>10591935
I typed that, and no for all x is no accident. It's saying that if you take any real number c and evaluate f and g at c and they turn out to be equal, then h' evaluated at c is 0. (I replaced x with c because it seems fixing x confuses you)

[Anonymous No.10591957]

>>10591952
>>10591938
Again, under the conditions that f(x)=g(x)

[Anonymous No.10591958]

>>10591919
I'm not who you're replying to, but you're fucking retarded lol His solution is completely right

[Anonymous No.10591959]

>>10591952
Try graphing h for f(x) = (x+1)^2, g(x) = (x+2)^2. And what f and g did you choose to get a constant h?

[Anonymous No.10591961]

>>10591958
So you also think h is a constant function?

[Anonymous No.10591969]

>>10591954
Cool, so then my reply to that post was correct except for that typo

[Anonymous No.10591978]

>>10591969
h is not a constant function, have you tried graphing it?

[Anonymous No.10591985]

>>10591969
No, you have shown that h(x)=(-0.5(a+b)+a)^2+(-0.5(a+b)+b)^2 when x=-0.5(a+b)
That doesn't tell you anything about h'(x)

[Anonymous No.10591993]

>>10591985
It tells me that h'(x) is 0

[Anonymous No.10591997]

>>10591993
Why doesn't id(x)=0 when x=0 tell me that id'(x)=0 then?

[Anonymous No.10591999]

>>10591997
Because that is for a specific x whereas mine is for all x which satisfy the conditions

[Anonymous No.10592002]

>>10591999
It's not, yours is specifically for x=-0.5(a+b). Have you tried graphing h yet?

[Anonymous No.10592007]

>>10591999
The condition your x satisfies is f(x) = g(x), which is precisely when x=-0.5(a+b) as you have deduced.

[Anonymous No.10592010]

>>10592002
That is the only valid x. The resulting function has a constant value

[Anonymous No.10592013]

>>10592007
Correct

[Anonymous No.10592020]

>>10592013
Wrong.

[Anonymous No.10592023]

>>10592020
No, that statement was undoubtedly correct

[Anonymous No.10592038]

>>10592023
Fuck you

[Anonymous No.10592044]

>>10592038
I'm into it, but this is a worksafe board

[Anonymous No.10592047]

>>10592044
Are you a cute grill or trap?

[Anonymous No.10592055]

>>10592047
No

[Anonymous No.10592058]

>>10592055
Faggot

[Anonymous No.10592072]

>>10592058
You too. Dibs on first top BTW. You can work my ass when I'm done with you

[Anonymous No.10592075]

>>10592072
Are you skinny? Have your considered hrt?

[Anonymous No.10592083]

>>10592058
>Faggot
Why the homophobia?

[Anonymous No.10592095]

>>10592075
Not skinny in a feminine way and I don't have any self confidence issues so no to considering hrt. Have you considered how you're going to make the journey to Cincinnati?

[Anonymous No.10592098]

>>10592095
What journey to Cincinnati?

[Anonymous No.10592103]

>>10592095
>I don't have any self confidence issues so no to considering hrt.
Don't you want to be a cute girl?

[Anonymous No.10592108]

>>10592098
The one you need to take as the initiator if you want this to happen

[Anonymous No.10592111]

>>10592108
I'm not gay. I'd do it if you take hrt tho

[Anonymous No.10592112]

>>10592111
>Not gay
>Would have sex with a man
I've got news for you, pally

[Anonymous No.10592117]

>>10592112
Do you know what hrt is?

[Anonymous No.10592119]

>>10592117
Hormone replacement therapy. For men it increases estrogen and decreases testosterone. They're still men unless you want to be pedantic and they're also trans, but that would make you queer

[Anonymous No.10592124]

>>10592119
Traps are not gay.

[Anonymous No.10592134]

>>10592124
Traps are gay by definition. If you like to have sex with traps or trans then you are at least queer. Gay and bisexual are also on the table. So which do you like best, creampuff?

[Anonymous No.10592138]

>>10592134
>Traps are gay by definition.
Wrong.
>If you like to have sex with traps or trans then you are at least queer. Gay and bisexual are also on the table. So which do you like best, creampuff?
You wouldn't suck Astolfo's peepee and swallow his seeds?

[Anonymous No.10592145]

>>10592138
>Wrong.
I don't think you know what traps are

>You wouldn't suck Astolfo's peepee and swallow his seeds?
Sure, just like how I'd cream your anonymous ass. That's how bisexuality works, the parts aren't super important

[Anonymous No.10592146]

>>10592145
>I don't think you know what traps are
Traps are not gay because they look like girls.

[Anonymous No.10592150]

>>10592146
Traps are gay because they're men who want to have sex with men and you are queer because you want to have sex with men dressed like women. It's 2019, get into it

[Anonymous No.10592159]

>>10592150
Whatever, faggot

[Anonymous No.10592161]

>>10592159
Enjoy the closet, queer

[Anonymous No.10592166]

(x+a)^2 = (x+b)^2.
2ax + a^2 = 2bx + b^2.
a^2 + 2x(a-b) + b^2 = 0.
d/dx(a^2 + 2x(a-b) + b^2) = 2(a-b).

true IF a and b are the same signs, otherwise the theorem fails.
example: with a,b = 1,-1: f(0) = g(0). d/dx{(x+1)^2 + (x-1)^2} = 4x = 0.
a,b = 1: f(n) = g(n) for all n. d/dx(2(x+1)^2) = 4(x+1) = 4.

[Anonymous No.10592177]

[eqn]\frac d{dx} 2x^2 + 2x(a+b) + a^2 + b^2 = 2(x +(a+b))[/eqn]
[eqn]g(x) = f(x) ~\text{when} ~a = \pm b.[/eqn]
[eqn] \text{when } a=b, f(x) = g(x) \text{ for all real numbers, but the derivative is only 0 when x = a+b, so the theorem is conditionally true if a=b.}[/eqn]
[eqn] \text{when } a=-b, f(x) = g(x) \text{ at x=0, where the derivative is 0. unconditionally true if } a = -b[/eqn]

[Anonymous No.10592286]

>>10591216
Do your own homework faggot

[Anonymous No.10592319]

>>10591216
So, consider me retarded, but are h(x) and h'(x) equal, or not?
If they are, then a=b, h(x)=2a, and a=0

[Anonymous No.10592409]

>>10591216
7

[Anonymous No.10592549]

>>10591216
h(x) = 2f(x) = 2x^2 + 4xa + 4a^a
let
z(x) = 2x^2
c(x) = 4xa
v(x) = 4a^2
than
h(x) = z(x) + c(x) + v(x)
so
h'(x) = z'(x) + c'(x) + v'(x)
z'(x) = 3x
c(x) = 0
v(x) = 0
h'(x) = 3x + 2
therefore its false

[Anonymous No.10592677]

>>10591216
>>10591741
FINDING X SUCH THAT g(x) = f(x):
g(x) = f(x)
(x+a)^2 = (x+b)^2
x+a = -x-b
x = (-a-b)/2

EVALUATING h'(x) AT THAT X:
h(x) = (x+a)^2 + (x+b)^2
h'(x) = 2(x+a) + 2(x+b) = 4x + 2(a+b)
h'((-a-b)/2) = 4(-a-b)/2 + 2(a+b) = 0

easy