\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
If we replace \(x\) with \(-x\) in the original equation, we get \[ f(-x) + (1 - (-x))f(-(-x)) = (-x)^2, \] which simplifies to \[ f(-x) + (1 + x)f(x) = x^2 \] as desired.
Therefore, we have a pair of equations in terms of \(f(x)\) and \(f(-x)\): \[ \begin {cases} f(x) + (1-x) f(-x) & = x^2 \\ (1+x) f(x) + f(-x) & = x^2. \end {cases} \]
Multiplying the second equation by \((1-x)\) gives us \[ (1-x^2)f(x) + (1-x)f(-x) = x^2(1-x), \] and subtracting the first equation from this \[ -x^2 f(x) = -x^3, \] which gives \(f(x) = x\).
Plugging this back, we have \begin {align*} \LHS & = f(x) + (1-x) f(-x) \\ & = x + (1-x) (-x) \\ & = x - x + x^2 \\ & = x^2 \\ & = \RHS \end {align*}
which holds. Therefore, \(f(x) = x\) is the solution to the functional equation.
For \(x \neq 1\), we have \begin {align*} K(K(x)) & = \frac {\frac {x + 1}{x - 1} + 1}{\frac {x + 1}{x - 1} - 1} \\ & = \frac {(x + 1) + (x - 1)}{(x + 1) - (x - 1)} \\ & = \frac {2x}{2} \\ & = x, \end {align*}
for \(x \neq 1\), as desired.
The equation on \(g\) is \[ g(x) + x g(K(x)) = x, \] and if we substitute \(x\) as \(K(x)\), we have \[ g(K(x)) + K(x) g(K(K(x))) = K(x), \] which simplifies to \[ g(K(x)) + K(x) g(x) = K(x). \]
Multiplying the second equation by \(x\), we have \[ xK(x) g(X) + x g(K(x)) = x K(x), \] and subtracting the first equation from this gives \[ (x K(x) - 1) g(x) = x (K(x) - 1), \] which gives \begin {align*} g(x) & = \frac {x \left (K(x) - 1\right )}{x K(x) - 1} \\ & = \frac {x \left (\frac {x + 1}{x - 1} - 1\right )}{x \cdot \frac {x + 1}{x - 1} - 1} \\ & = \frac {x \left [(x + 1) - (x - 1)\right ]}{x (x + 1) - (x - 1)} \\ & = \frac {2x}{x^2 + 1}, \end {align*}
for \(x \neq 1\).
If we plug this back to the original equation, we have \begin {align*} \LHS & = \frac {2x}{x^2 + 1} + x \frac {2 \cdot \frac {x + 1}{x - 1}}{\left (\frac {x + 1}{x - 1}\right )^2 + 1} \\ & = \frac {2x}{x^2 + 1} + \frac {2x \cdot (x + 1) \cdot (x - 1)}{(x + 1)^2 + (x - 1)^2} \\ & = \frac {2x}{x^2 + 1} + \frac {2x(x^2 - 1)}{2x^2 + 2} \\ & = \frac {2x}{x^2 + 1} + \frac {x(x^2 - 1)}{x^2 + 1} \\ & = \frac {x^3 - x + 2x}{x^2 + 1} \\ & = \frac {x(x^2 + 1)}{x^2 + 1} \\ & = x \\ & = \RHS , \end {align*}
so \[ g(x) = \frac {2x}{x^2 + 1} \] is the solution to the original functional equation.
Let \(H(x) = \frac {1}{1 - x}\). Notice that \begin {align*} H(H(x)) & = \frac {1}{1 - \frac {1}{1 - x}} \\ & = \frac {1 - x}{1 - x - 1} \\ & = \frac {x - 1}{x} \\ & = 1 - \frac {1}{x} \end {align*}
and \begin {align*} H(H(H(x))) & = \frac {1}{1 - \left (1 - \frac {1}{x}\right )} \\ & = \frac {x}{1} \\ & = x. \end {align*}
Now, if we replace all the \(x\) with \(\frac {1}{1 - x}\), we will get \[ h\left (\frac {1}{1 - x}\right ) + h\left (1 - \frac {1}{x}\right ) = 1 - \frac {1}{1 - x} - \left (1 - \frac {1}{x}\right ), \] and doing the same replacement again gives us \[ h\left (1 - \frac {1}{x}\right ) + h(x) = 1 - \left (1 - \frac {1}{x}\right ) - x. \]
Summing these two equations, together with the original equation, gives us that \[ 2 \cdot \left [h\left (\frac {1}{1 - x}\right ) + h\left (1 - \frac {1}{x}\right ) + h(x)\right ] = 3 - 2 \cdot \left [x + \frac {1}{1 - x} + \left (1 - \frac {1}{x}\right )\right ], \] and therefore \[ h\left (\frac {1}{1 - x}\right ) + h\left (1 - \frac {1}{x}\right ) + h(x) = \frac {3}{2} - \left [x + \frac {1}{1 - x} + \left (1 - \frac {1}{x}\right )\right ]. \]
Subtracting the second equation from this, gives that \begin {align*} h(x) & = \left (\frac {3}{2} - \left [x + \frac {1}{1 - x} + \left (1 - \frac {1}{x}\right )\right ]\right ) - \left [1 - \frac {1}{1 - x} - \left (1 - \frac {1}{x}\right )\right ] \\ & =\frac {1}{2} - x. \end {align*}
Plugging this back to the original equation, we have \begin {align*} \LHS & = \frac {1}{2} - x + \frac {1}{2} - \frac {1}{1 - x} \\ & = 1 - x - \frac {1}{1 - x} \\ & = \RHS , \end {align*}
which satisfies the original functional equation. Therefore, the original equation solves to \[ h(x) = \frac {1}{2} - x. \]