Solving For A Nice Function

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[cd_hnds]

Shouldn't the difference in domain be mentioned as well as part of the solution?

[RajAgarwal-pgkn]

I think you can generalise the second method just a little...

Say f((ax + b)/(cx + d)) = ((cx + d)/(lx + m))

Then we follow the methodogy of taking the reciprocal as you did, and then break the denominator as follows

n(ax +b) + p(cx + d) = lx + m

solve for n and p [it wont always haave a solution, but it still as formalised general explanation for the methodoly], split the denominator into the sum of the whole number p and partial fraction (n(ax + b)/(cx + d)), then substitute the partial fraction with another variable (what you called y), and get the solution.

[tenkey]

I substituted x by (λ+1), i got f((λ+2)/λ)= λ/(4λ+2), then i substitued λ again by 1/x . after simplifying, i found : f(2x+1) = 1/(2x+4) and then i found that f(x) = 1/(x+3) for all real x not equal to -3, 1/2 and 1

[henricovsky]

Why did you substitute x with x+1/x-1 just to get f(y) if you had already called x+1/x-1 = y? Really love the videos though

[bohemiathakur]

how do you make these questions, i want to learn

[bobmarley]

it's also interesting to note how the input x into f((x+1)/(x-1)) can't be 1 b/c of zero denominator, but also how f(x) can't have input x being 1 as well, since (x+1)/(x-1) can never equal 1... however in the f(x) you obtained 1/(x+3), it makes it seem that you can input x = 1 into f(x).... so I'm a little confused here and would need some clarification as to what is the correct answer? Kindly please let me know :)

[anestismoutafidis]

Definition of quantity=D
D={ℝ\ -1; 1; 0, 5}
f(x+1/x-1) - (x-1/4x-2)=0
f(0+1/0-1) - (0-1/4•0 -2)= -1 - 1/2= -1, 5
x≠0; <=> f• (x-1/4x-2) =>

(x+1)/(4x-2)= |0, 45| +1/4•|0, 45|-2
<=>1, 45/3, 8=0, 38 <=> x=|0, 45l
We make the proof with functional constants:
[|ln(π•5, 5φ/e^2)|]+1÷[|ln(π•5, 5φ/e^2)|]-1=
[|ln(π•5, 5φ/e^2)|]-1÷(4•[|ln(π•5, 5φ/e^2|)]-2
<=>[ 0, 38 = 0, 38 with x=|0, 45|]
x=|0, 45| = |ln(π•5, 5φ)/e^2|
Functional operator: (x-1) with limit 0, 38^2