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.
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.
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.
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
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.
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.
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.
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
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.
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.
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
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.
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
nice
As a kindergarten prodigy, hi
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.
peak
If f((3x)/(2x+1))=4/(5x-6),then f(x)=(12-8x)/(-18+17x) final answer
I think this is grade 9 or 10
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 👌
the question is why. Never seen that type of function in calculus, so pointless to solve.