2:29 you can just plug in t since you've already defined it in the Let statement

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.

As a 10th grader, I have no idea why I am watching these videos but they are just super addicting

Cool!
I don't think I've ever seen this before.

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.

I got a similar question in Math Olympics I did almost this but didn't know how to continue

Set y=(ax+b)/(cx+d) and subsititute for x in f(x) and set that equal to 4/(5x-6). remove parenthesis and bring everything to one side. From there you derive three linear equations in four letters where you can find any combo for a,b,c,d that works. Not necessarily quicker, just different

Your teaching style and solution method are great.👏

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)

You are a genius!!! It's the true. Thank you very much.

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 was thinking i hard trying to understanf. Just to realisd is just composite function.

Which part of a precalculus (algebra & trigonometry) textbook is that kind to stuff usually covered?  Composite function?  The algorithm demonstrated to decompose f(x) from f(g(x)) is beautiful.

Damn these are addicting. I binge watched till the end for the satisfying answer.

Can you suggest a book where I can found a chapter dedicated to functional equations and exercises?

Hm substitution, which indirectly requires inverse function yes. This is a universal trick, except that not all functions have inverse functions, especially not mapping R bijectively.
So u := 3x/(2x+1) -> trouble ->
Ansatz c*(u+a)/(u+b)
= c*(3x + a(2x+1)) / (3x + b(2x+1))  this should deliver x, find a, b, c:
Denominator = const, means b=-3/2
nominator is better without plus const, so a = 0
Having c * 3x/(-3/2) =: x, so c = -1/2
Having -1/2 * u / (u - 3/2) == x
Replace right-hand-side, no hurry for simplification, gets f(u) = expression in u.
Then sneakingly rename u to x again 😅

Generalization f(g(x)) = h(x) => y = g(x) so x = g^inv(y) = G(y), so f(y) = h(G(y)) which can then be re-written as f(x) = h(G(x))
Here, y = g(x) = 3x / (2x + 1), so 1/y = 2/3 + 1 / 3x, so x = 1 / (3/y - 2) = G(y), so h(x) = 4 / (5x - 6) = 4 / (5G(y) - 6) = 4 / [ 5 / (3/y - 2) - 6] = 4 / {[ 5y - 6 (3-2y)] / (3-2y)} = (12 - 8y) / (17y + 18)
Thus, f(x) = h(G(x)) = (12 - 8x) / (17x + 18)

Boas festas e muita saúde professor! 👍🙏🙏🙏

What do you use such equations for in practical terms?

Damg, Ik this is simple but it still excites me.