Yeah. I kept waiting for him to use the phrase, "now let's take this equation from the X world to the T world." I mean, this is basically a U-sub solve.
@@somehand986 at the time stamp, he substitutes t/(3-2t) for x into f(3x/(2x+1)) and then reduces it down to f(t). This was an unnecessary calculation, as he had previously defined 3x/(2x+1)=t, so he could have saved a few steps by substituting that strait away to get f(3x/(2x+1))=f(t) without additional work, and then continue with the solution as he did.
But I think I get why he did it this way; it can be unintuitive to substitute an expression with a variable, depending on experience. But yeah, he should have pointed out the other, cleaner, substitution.
We don’t even need to substitute x with the expression for t inside the function because we already said in the very first line that 3x/(2x+1)=t so we know the LHS is f(t) straight away. Still great video.
A few weeks ago, I noticed something interesting about inverses of rational functions of the form f(x)=(Ax+B)/(Cx+D) and 2x2 matrices: finv(x) = (Dx-B)/(-Cx+A) = (D/(AD-BC)x-B/(AD-BC))/(-C/(AD-BC)x+A/(AD-BC)) Notice the similarity to the inverse of the 2x2 matrix (A B);(C D) is (D/(AD-BC) -B/(AD-BC));(-C/(AD-BC). A/(AD-BC)) I’m not the first person to notice this but I can’t find a satisfactory explanation for this that goes beyond “well they’re both inverses”. I’d really like to understand if there’s a deeper connection that goes beyond interesting coincidence. Otherwise, it’s just a nice mnemonic shortcut to finding the inverse of a linear/linear rational function.
Given that bprp used the term inverse function, perhaps it would have been conceptually clearer to spell out the composition of functions and say that f(g(x))=4/(5x-6), with g(x) =3x/(2x+1) as the function to be "inverted". So setting t=g(x), the search is for x=g_inverse(t) to substitute into the original RHS.
There's a "trick" to this type of question. (there are reasons behind this and it's not a coincidence) Any function of the form f(x) = (ax+b)/(cx+d) can be thought of as a 2x2 matrix like this: [a, b] [c, d] To get the inverse function, you just invert this matrix. If D is the determinant, then the inverse matrix is: [d/D, -b/D] [-c/D, a/D] So the inverse of f is (dx/D -b/D)/(-cx/D + a/D), which you can simplify further by multiplying the numerator and denominator by D to get (dx -b)/(-cx + a), which is much cleaner. (note that the case when D = 0, which would prevent us from dividing by D, is really just the case when the top row is a multiple of the bottom row, and the entire function would simplify to a constant, so it never arises) If you want to compose 2 such functions, you would multiply their matrices, then take that matrix and reinterpret it back as a function. In the case of this question, you have g(x) = 3x/(2x+1) and h(x) = 4/(5x-6), and you're given that f(g(x)) = h(x). If we let k(x) be the inverse of g, then to get f(x), you need to substitute x with k(x) and get f(x) = h(k(x)) Apply the above to the question, and you get that the matrix for k is [1,0] [-2,3] and the matrix for h is [0,4] [5,-6] and then you multiply them (in the correct order, which is h*k) to get [-8, 12] [17, -18] and this is the function f(x) = (-8x+12)/(17x-18), just like the video.
You don’t necessarily need to get involved with matrices though. You can use just the rule derived from it. If f(x)=(ax+b)/(cx+d) then f^-1(x)=(-dx+b)(cx-a) Define the inside of f() of the left side as g(x) and find its inverse. Which is g^-1(x)=(-x)/(2x-3) Plug this instead of x on the other side of the equation and you get the answer (8x-12)/(-17x+18) (Bprp’s answer is the same multiplied by -1 thats why it looks different)
@@f5673-t1h I mean if we suppose you don't know matrices this is the easiest solution that came to my mind. A lot more straightforward than bprp's solution that's for sure.
Both simply represent independent variable - you can call it x, or t (which is often used for time dependent functions), or z or whatever you like. You are not changing the function, just the *name* of the input variable.
@@LetsTaIk In problems like this, when you have just f(t) on one side of the equation and the resulting function on the other (e.g. [x+3]/[2x+5]), 't' is what we call a 'dummy variable'. What this means is that we can replace it with any new, unrelated variable we want. For example, we could use 'u', 'w' or even 'p' if we wanted to. Your problem with the question relates to a small but significant part of what I said above: the variable should be unrelated to the question in any previous algebra done. You are completely correct that 'x' shouldn't be used here, since it was defined beforehand in terms of 't'. However, for the sake of simplicity, and since functions are normally defined in terms of 'x', he replaced 't' with 'x' in the video. I hope this was helpful. If not, I'm sure that others in this comment section will do a better job in the future.
Very straightforward. But still, no need to make the complicated calculation in the second part of the solution which is prone to errors. Since we said let t=3x/(2x+1) then once x(t) is found, We simply write: f(t)=4/(5x(t)-6) and replace x(t) by its value in terms of t.
I would like some explanations If x=0 we have f(0) = F(0) and f(0)= F(0) = -4\6 For x=1 we have f(1)=F(1)= -4 ok but if x=2 we have f(6\5) and f(6\5) does not exist but F(6\5)=-4\11
here's a funny problem checking your logarithm and induction skills that i might as well share: log2(3) * log3(4) * log4(5) * log5(6)... * log1022(1023) * log1023(1024) = ? some might already see the solution by seeing the numbers available
There is no t^2 term. When he carried out the multiplication in the numerator and denominator by (3 - 2t), he went [5t/(3 - 2t)] * (3 - 2t), which cancels out to leave only 5t for that term. Then multiplying -6 by (3 - 2t) gives -18 + 12t for the second term. Adding the terms gives 5t - 18 + 12t, or 17t - 18.
Here is a flaw: Your result of f(x)=whatever has domain on R which includes x=1.5 However the precondition f(3x/(2x+1))=whatever the 3x/(2x+1) can NEVER be 1.5 this is the characteristics of inverse proportional function. (it's 1.5-1.5/(2x+1) which the division won't end in 0). So the precondition really tells you NOTHING about what equals f(1.5). What is the answer then?? is it undefined or not exist or what?
I think we just assume that we are working inside the domain where (let's call (3x)/(2x+1) g(x)) g(x) is also defined which is just another function. As for a solution If f(x)=(ax+b)/(cx+d) then f^-1(x)=(-dx+b) (cx-a) So we find the inverse of g(x) which is g^-1(x)=(-x)/(2x-3) Plug this instead of x on the other side of the equation and you get the answer (8x-12)/(-17x+18) (Bprp's answer is the same with both terms multiplied by -1 thats why it looks different)
@@SegFaultOnLine1984 You didn't understand. I was not questioning the main part of the answer. Since the precondition didn't tell you what f(1.5) is you CAN'T say you got a solution for f(1.5).
@phoenixarian8513 The fact that the "precondition" or original information doesn't completely cover the full real line could be an issue in principle - e.g. in a case where there was no information about negative arguments, but I think that here continuity is sufficient to address your concern. In the limit where the original x->infinity (either direction) and argument of f() -> 1.5, the RHS -> zero, and that's exactly what BPRP's solution gives for f(3/2). You ought to at least admit it is consistent. Similiarly, the original information shows (consider x->-1/2) that f(t) -> -8/17 as t-> infinity (both directions) which is again in agreement.
I'm sorry but there is a mistake :) Moreover there is a shorter way, so beautiful... :! I love what you do, and every day I get better at math thanks to you ! f(3x/2x+1) = 4/(5x-6) = f(3x)/(2x+1) then f (x) = 4 (2x+1) / 3 (5x-6) Thank's ! :)
As a 10th grader, I have no idea why I am watching these videos but they are just super addicting
pookie someone from middle school probably is watching this
I'm in 11th and it is useful
Im from kindergarten my little pookie cookie@@jir_UwU
Bro I am from university and still don't get it
but they are addictive 😅
I’m in tenth grade watching it and I’m in precalculus lol
2:29 you can just plug in t since you've already defined it in the Let statement
Yeah. I kept waiting for him to use the phrase, "now let's take this equation from the X world to the T world." I mean, this is basically a U-sub solve.
Could you explain it😅? I don’t get it
@@somehand986 at the time stamp, he substitutes t/(3-2t) for x into f(3x/(2x+1)) and then reduces it down to f(t). This was an unnecessary calculation, as he had previously defined 3x/(2x+1)=t, so he could have saved a few steps by substituting that strait away to get f(3x/(2x+1))=f(t) without additional work, and then continue with the solution as he did.
But I think I get why he did it this way; it can be unintuitive to substitute an expression with a variable, depending on experience. But yeah, he should have pointed out the other, cleaner, substitution.
Exactly ! what is he doing ? He defined t as being 3X/(2X+1), why the hell does he go recalculating it 🤯
We don’t even need to substitute x with the expression for t inside the function because we already said in the very first line that 3x/(2x+1)=t so we know the LHS is f(t) straight away.
Still great video.
Your teaching style and solution method are great.👏
A few weeks ago, I noticed something interesting about inverses of rational functions of the form f(x)=(Ax+B)/(Cx+D) and 2x2 matrices:
finv(x) = (Dx-B)/(-Cx+A) = (D/(AD-BC)x-B/(AD-BC))/(-C/(AD-BC)x+A/(AD-BC))
Notice the similarity to the inverse of the 2x2 matrix (A B);(C D) is (D/(AD-BC) -B/(AD-BC));(-C/(AD-BC). A/(AD-BC))
I’m not the first person to notice this but I can’t find a satisfactory explanation for this that goes beyond “well they’re both inverses”. I’d really like to understand if there’s a deeper connection that goes beyond interesting coincidence. Otherwise, it’s just a nice mnemonic shortcut to finding the inverse of a linear/linear rational function.
ruclips.net/video/hG8pIuVYZEg/видео.htmlsi=FFIVkziX-LeQvySs
f[3x/(2x +1)] = 4/(5x -6) find f(x):
Let y = 3x/(2x +1) ⇒ 2xy +y -3x = 0
⇒ x (2y -3) = -y ⇒ x = y/(3 -2y)
⇒ f(y) = 4/[5y/(3 -2y) -6] = 4 (3 -2y)/(17y -18)
⇒ f(x) = (12 -8x)/(17x -18)
What about the valid domain can’t x not be -0.5 or 1.2 in initial functional equation but it would be ok in the new one?
Given that bprp used the term inverse function, perhaps it would have been conceptually clearer to spell out the composition of functions and say that f(g(x))=4/(5x-6), with g(x) =3x/(2x+1) as the function to be "inverted". So setting t=g(x), the search is for x=g_inverse(t) to substitute into the original RHS.
Can you suggest a book where I can found a chapter dedicated to functional equations and exercises?
There's a "trick" to this type of question. (there are reasons behind this and it's not a coincidence)
Any function of the form f(x) = (ax+b)/(cx+d) can be thought of as a 2x2 matrix like this:
[a, b]
[c, d]
To get the inverse function, you just invert this matrix. If D is the determinant, then the inverse matrix is:
[d/D, -b/D]
[-c/D, a/D]
So the inverse of f is (dx/D -b/D)/(-cx/D + a/D), which you can simplify further by multiplying the numerator and denominator by D to get (dx -b)/(-cx + a), which is much cleaner.
(note that the case when D = 0, which would prevent us from dividing by D, is really just the case when the top row is a multiple of the bottom row, and the entire function would simplify to a constant, so it never arises)
If you want to compose 2 such functions, you would multiply their matrices, then take that matrix and reinterpret it back as a function.
In the case of this question, you have g(x) = 3x/(2x+1) and h(x) = 4/(5x-6), and you're given that f(g(x)) = h(x). If we let k(x) be the inverse of g, then to get f(x), you need to substitute x with k(x) and get f(x) = h(k(x))
Apply the above to the question, and you get that the matrix for k is
[1,0]
[-2,3]
and the matrix for h is
[0,4]
[5,-6]
and then you multiply them (in the correct order, which is h*k) to get
[-8, 12]
[17, -18]
and this is the function f(x) = (-8x+12)/(17x-18), just like the video.
You don’t necessarily need to get involved with matrices though. You can use just the rule derived from it.
If f(x)=(ax+b)/(cx+d) then f^-1(x)=(-dx+b)(cx-a)
Define the inside of f() of the left side as g(x) and find its inverse. Which is g^-1(x)=(-x)/(2x-3)
Plug this instead of x on the other side of the equation and you get the answer (8x-12)/(-17x+18)
(Bprp’s answer is the same multiplied by -1 thats why it looks different)
@@SegFaultOnLine1984 Yes, you can plug it in, but people generally don't want to deal with clearing denominators and that messiness.
@@f5673-t1h I mean if we suppose you don't know matrices this is the easiest solution that came to my mind. A lot more straightforward than bprp's solution that's for sure.
You are a genius!!! It's the true. Thank you very much.
Cool!
I don't think I've ever seen this before.
why can you replace t with x in the last step?
Both simply represent independent variable - you can call it x, or t (which is often used for time dependent functions), or z or whatever you like. You are not changing the function, just the *name* of the input variable.
@@mepersonexcept we said t is equal to (3x)/(2x+1) earlier, not t = x. I don’t understand.
@@LetsTaIk In problems like this, when you have just f(t) on one side of the equation and the resulting function on the other (e.g. [x+3]/[2x+5]), 't' is what we call a 'dummy variable'. What this means is that we can replace it with any new, unrelated variable we want. For example, we could use 'u', 'w' or even 'p' if we wanted to.
Your problem with the question relates to a small but significant part of what I said above: the variable should be unrelated to the question in any previous algebra done. You are completely correct that 'x' shouldn't be used here, since it was defined beforehand in terms of 't'. However, for the sake of simplicity, and since functions are normally defined in terms of 'x', he replaced 't' with 'x' in the video.
I hope this was helpful. If not, I'm sure that others in this comment section will do a better job in the future.
bc its same variable on both sides, doesnt matter, it could be a flower
@@meperson yeah but x isnt t, wouldnt it be f(t/(3-2t))=(12-8t)/(-18+17t)
Hi. Why do you specify "for precalculus students"? Is there another way to solve it with calculus?
i was thinking i hard trying to understanf. Just to realisd is just composite function.
4(2x + 1）/3(5x - 6)
Very straightforward.
But still, no need to make the complicated calculation in the second part of the solution which is prone to errors.
Since we said let t=3x/(2x+1)
then once x(t) is found,
We simply write:
f(t)=4/(5x(t)-6) and replace x(t) by its value in terms of t.
Boas festas e muita saúde professor! 👍🙏🙏🙏
Why you replace in the final t by x .
Non è importante il nome della variabile, quindi ha sostituito con x per rispettare la risposta f(x)=?
f(x) ha lo stesso grafico di f(t).
let y = 3x /(2x + 1)
y(2x +1) = 3x
x(3 - 2y) = y
x = y/(3 - 2y)
5x - 6 = 5y/(3 - 2y) - 64
= (5y - 6(3 - 2y))/(3 - 2y)
= (17y - 18)/(3 - 2y)
4/(5x - 6) = 4(3-2y)/(17y - 18)
f(x) = 4(3-2x)/(17x-18)
I would like some explanations If x=0 we have f(0) = F(0) and f(0)= F(0) = -4\6 For x=1 we have f(1)=F(1)= -4 ok but if x=2 we have f(6\5) and f(6\5) does not exist but F(6\5)=-4\11
Is this composition function
here's a funny problem checking your logarithm and induction skills that i might as well share:
log2(3) * log3(4) * log4(5) * log5(6)... * log1022(1023) * log1023(1024) = ?
some might already see the solution by seeing the numbers available
Ans is 10
log(1024)/log(2) (base ten) = 10. Great problem!
Telescopic product
how would you even solve this idk
@@stupidteous Everything cancels out and you're left with ukytr's calculation
I believe I have a more elegant way to to it.
1) replace x with 1/x
2) replace x with x-2
3) replace x with 3/x
Damg, Ik this is simple but it still excites me.
So easy look like find f_1(x)
Acertei 👍
Đặt t bằng cái f() rồi còn thế ra thế vô
Great
(3x)/(2x - 1) = y
(3x) = y(2x - 1)
y = x(2y - 3)
x = y/(2y - 3)
5x - 6 = 5y/(2y - 3) - 6 = (18 - 7y)/(2y - 3)
4/(5x - 6) = 4(2y - 3)/(18 - 7y)
f(x) = 4(2x - 3)/(18 - 7x) = (8x - 12)/(18 - 7x)
Why I will want to solve this ? What this use in life ?
If f((3x)/(2x+1))=4/(5x-6),then f(x)=(12-8x)/(-18+17x) final answer
nice
peak
wait... (5t/(3-2t) -6)x(3-2t)... where did the 12t square go?
There is no t^2 term.
When he carried out the multiplication in the numerator and denominator by (3 - 2t), he went [5t/(3 - 2t)] * (3 - 2t), which cancels out to leave only 5t for that term.
Then multiplying -6 by (3 - 2t) gives -18 + 12t for the second term. Adding the terms gives 5t - 18 + 12t, or 17t - 18.
@@pikespeakaudio8898 👌
I feel like you really over complicated this one
that's the most intuitive solution except 2:29 where you can just plug in t, like f(t)
@@alibekturashev6251i think i found a more intuitive solution. Define the inside of f() as g(x) and find its inverse. Then just plug it in the right.
how bout you solve it then 🤥🤥🤥
Exactly
As a kindergarten prodigy, hi
Here is a flaw: Your result of f(x)=whatever has domain on R which includes x=1.5
However the precondition f(3x/(2x+1))=whatever the 3x/(2x+1) can NEVER be 1.5 this is the characteristics of inverse proportional function. (it's 1.5-1.5/(2x+1) which the division won't end in 0). So the precondition really tells you NOTHING about what equals f(1.5).
What is the answer then?? is it undefined or not exist or what?
I think we just assume that we are working inside the domain where (let's call (3x)/(2x+1) g(x)) g(x) is also defined which is just another function.
As for a solution If f(x)=(ax+b)/(cx+d) then f^-1(x)=(-dx+b) (cx-a)
So we find the inverse of g(x) which is g^-1(x)=(-x)/(2x-3)
Plug this instead of x on the other side of the equation and you get the answer (8x-12)/(-17x+18)
(Bprp's answer is the same with both terms multiplied by -1 thats why it looks different)
@@SegFaultOnLine1984 You didn't understand. I was not questioning the main part of the answer. Since the precondition didn't tell you what f(1.5) is you CAN'T say you got a solution for f(1.5).
@phoenixarian8513 The fact that the "precondition" or original information doesn't completely cover the full real line could be an issue in principle - e.g. in a case where there was no information about negative arguments, but I think that here continuity is sufficient to address your concern. In the limit where the original x->infinity (either direction) and argument of f() -> 1.5, the RHS -> zero, and that's exactly what BPRP's solution gives for f(3/2). You ought to at least admit it is consistent. Similiarly, the original information shows (consider x->-1/2) that f(t) -> -8/17 as t-> infinity (both directions) which is again in agreement.
Messy explanation. 😳
I'm sorry but there is a mistake :) Moreover there is a shorter way, so beautiful... :! I love what you do, and every day I get better at math thanks to you ! f(3x/2x+1) = 4/(5x-6) = f(3x)/(2x+1) then f (x) = 4 (2x+1) / 3 (5x-6) Thank's ! :)
you can't say that f(3x/(2x+1)) = f(3x)/(2x+1). he didn't do anything wrong
Wrong
Basic understanding is wrong
the question is why. Never seen that type of function in calculus, so pointless to solve.