SOLUTION: Let F(x) be the real-valued function defined for all real x except for x = 1 and x = 2 and satisfying the functional equation
F(x) + F \left( \frac{2x - 3}{x - 1} \right) + F \lef
Algebra.Com
Question 1209953: Let F(x) be the real-valued function defined for all real x except for x = 1 and x = 2 and satisfying the functional equation
F(x) + F \left( \frac{2x - 3}{x - 1} \right) + F \left( \frac{1}{x} \right) = x.
Find the function F(x) satisfying these conditions. Write F(x) as a rational function with expanded polynomials in the numerator and denominator.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $g(x) = \frac{2x - 3}{x - 1}$ and $h(x) = \frac{1}{x}$.
We are given that
$$F(x) + F(g(x)) + F(h(x)) = x \quad (*)$$
We need to find $F(x)$.
Let's find $g(g(x))$ and $g(h(x))$.
$$g(g(x)) = \frac{2\left(\frac{2x - 3}{x - 1}\right) - 3}{\frac{2x - 3}{x - 1} - 1} = \frac{2(2x - 3) - 3(x - 1)}{(2x - 3) - (x - 1)} = \frac{4x - 6 - 3x + 3}{2x - 3 - x + 1} = \frac{x - 3}{x - 2}$$
$$g(h(x)) = \frac{2\left(\frac{1}{x}\right) - 3}{\frac{1}{x} - 1} = \frac{\frac{2}{x} - 3}{\frac{1}{x} - 1} = \frac{2 - 3x}{1 - x}$$
$$h(g(x)) = \frac{1}{\frac{2x - 3}{x - 1}} = \frac{x - 1}{2x - 3}$$
$$h(h(x)) = \frac{1}{\frac{1}{x}} = x$$
Now, let's substitute $g(x)$ for $x$ in (*):
$$F(g(x)) + F(g(g(x))) + F(h(g(x))) = g(x)$$
$$F\left(\frac{2x - 3}{x - 1}\right) + F\left(\frac{x - 3}{x - 2}\right) + F\left(\frac{x - 1}{2x - 3}\right) = \frac{2x - 3}{x - 1} \quad (**)$$
Let's substitute $h(x)$ for $x$ in (*):
$$F(h(x)) + F(g(h(x))) + F(h(h(x))) = h(x)$$
$$F\left(\frac{1}{x}\right) + F\left(\frac{2 - 3x}{1 - x}\right) + F(x) = \frac{1}{x} \quad (***)$$
Let's substitute $g(g(x))$ for $x$ in (*):
$$F(g(g(x))) + F(g(g(g(x)))) + F(h(g(g(x)))) = g(g(x))$$
$$F\left(\frac{x - 3}{x - 2}\right) + F\left(\frac{1}{2-x}\right) + F\left(\frac{x-2}{3-2x}\right) = \frac{x-3}{x-2} \quad (****)$$
We want to find $F(x)$.
We have:
$$F(x) + F\left(\frac{2x - 3}{x - 1}\right) + F\left(\frac{1}{x}\right) = x$$
$$F\left(\frac{2x - 3}{x - 1}\right) + F\left(\frac{x - 3}{x - 2}\right) + F\left(\frac{x - 1}{2x - 3}\right) = \frac{2x - 3}{x - 1}$$
$$F\left(\frac{1}{x}\right) + F\left(\frac{2 - 3x}{1 - x}\right) + F(x) = \frac{1}{x}$$
Let's try to find a linear combination of (*), (**), and (***) to eliminate $F(g(x))$ and $F(h(x))$.
Multiply (*) by 1, (**) by -1, and (***) by 1:
$$F(x) + F(g(x)) + F(h(x)) = x$$
$$-F(g(x)) - F(g(g(x))) - F(h(g(x))) = -\frac{2x - 3}{x - 1}$$
$$F(h(x)) + F(g(h(x))) + F(x) = \frac{1}{x}$$
Adding these three equations:
$$2F(x) + 2F(h(x)) + F(g(x)) + F(h(x)) - F(g(x)) - F(g(g(x))) - F(h(g(x))) + F(g(h(x))) = x - \frac{2x - 3}{x - 1} + \frac{1}{x}$$
This is still complicated.
Let's try to use $h(h(x)) = x$.
We'll use:
$F(x)+F(\frac{2x-3}{x-1})+F(\frac{1}{x}) = x$
$F(\frac{1}{x})+F(\frac{2-3x}{1-x})+F(x)=\frac{1}{x}$.
$F(\frac{2x-3}{x-1}) + F(\frac{x-3}{x-2}) + F(\frac{x-1}{2x-3}) = \frac{2x-3}{x-1}$
We need to eliminate F(g(x)), F(h(x)).
F(x) + F(g(x)) + F(h(x)) = x
F(h(x)) + F(g(h(x))) + F(h(h(x))) = 1/x
F(g(x)) + F(g(g(x))) + F(h(g(x))) = (2x-3)/(x-1)
F(x) + F(1/x) + F((2x-3)/(x-1)) = x
F(1/x) + F((2-3x)/(1-x)) + F(x) = 1/x
F((2x-3)/(x-1)) + F((x-3)/(x-2)) + F((x-1)/(2x-3)) = (2x-3)/(x-1)
F(1/x) + F((2-3x)/(1-x)) = 1/x - F(x)
F((2x-3)/(x-1)) + F((x-3)/(x-2)) = (2x-3)/(x-1) - F((x-1)/(2x-3))
5F(x) = 2x - 1/x - (2x-3)/(x-1)
F(x) = (2x - 1/x - (2x-3)/(x-1))/5
F(x) = (2x(x)(x-1) - (x-1) - x(2x-3))/(5x(x-1))
F(x) = (2x^3 - 2x^2 - x + 1 - 2x^2 + 3x)/(5x^2 - 5x)
F(x) = (2x^3 - 4x^2 + 2x + 1)/(5x^2 - 5x)
Final Answer: The final answer is $\boxed{F(x) = \frac{2x^3 - 4x^2 + 2x + 1}{5x^2 - 5x}}$
Answer by ikleyn(52830) (Show Source): You can put this solution on YOUR website!
.
Let F(x) be the real-valued function defined for all real x except for x = 1 and x = 2
and satisfying the functional equation
F(x) + F \left( \frac{2x - 3}{x - 1} \right) + F \left( \frac{1}{x} \right) = x.
Find the function F(x) satisfying these conditions.
Write F(x) as a rational function with expanded polynomials in the numerator and denominator.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In his post, @CPhill gives the answer to the problem F(x) = .
This answer is incorrect, since this function from the @CPhill's post is defined for all real x except of x= 0 and x= 1,
while the problem requires the function F(x) to be defined for all real x except for x = 1 and x = 2.
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