The function f satisfies, for all \(x\), the equation
\[
\f(x) + (1- x)\f(-x) = x^2\, .
\]
Show that \(\f(-x) + (1 + x)\f(x) = x^2\,\).
Hence find \(\f(x)\) in terms of \(x\). You should
verify that your function satisfies the original equation.
The function \({\rm K}\) is defined, for \(x\ne 1\), by
\[{\rm K}(x) = \dfrac{x+1}{x-1}\,.\]
Show that, for \(x\ne1\),
\({\rm K(K(}x)) =x\,\).
The function g satisfies the equation
\[
\g(x)+ x\, \g\Big(\frac{ x+1 }{x-1}\Big)
= x \ \ \ \ \ \ \ \ \ \ \
( x\ne 1)
\,.
\]
Show
that, for \(x\ne1\), \(\g(x)= \dfrac{2x}{x^2+1}\,\).
Find \(\h(x)\), for \(x\ne0\), \(x\ne1\), given that
\[
\h(x)+ \h\Big(\frac 1 {1-x}\Big)= 1-x -\frac1{1-x}
\ \ \ \ \ \ (
x\ne0, \ \ x\ne1 )
\,.
\]
Solution:
\(\,\) Let \(P(x)\) mean the proposition that \(f(x) + (1-x)f(-x) = x^2\) so
\begin{align*}
P(x): && f(x) + (1-x)f(-x) &= x^2 \\
P(-x): && f(-x)+(1+x)f(x) &= (-x)^2 = x^2 \\
\Rightarrow && f(x)+(1-x)\left (x^2-(1+x)f(x) \right) &= x^2 \\
\Rightarrow && f(x) \left (1 -(1-x^2) \right) &= x^2 + (x-1)x^2 \\
\Rightarrow && f(x)x^2 &= x^3 \\
\Rightarrow && f(x) &= x
\end{align*}
Notice that \(x + (1-x)(-x) = x^2\) so it does satisfy the functional equation.
Let \(K(x) = \frac{x+1}{x-1}\) if \(x \neq 1\) so
\begin{align*}
&& K(K(x)) &= \frac{K(x)+1}{K(x)-1} \\
&&&= \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \\
&&&= \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} \\
&&&= x
\end{align*}
Let \(Q(x)\) denote the proposition that \(g(x) + xg(K(x)) = x\) so
\begin{align*}
Q(x): && g(x) + xg(K(x)) &= x \\
Q(K(x)): && g(K(x)) + K(x)g(x) &= K(x) \\
\Rightarrow && g(x) +xK(x)[1-g(x)] &= x \\
\Rightarrow && g(x)[1-xK(x)] &= x(1-K(x)) \\
\Rightarrow && g(x) \frac{x-1-x^2-x}{x-1} &= \frac{-2x}{x-1} \\
\Rightarrow && g(x) &= \frac{2x}{x^2+1}
\end{align*}.
And notice that \(\frac{2x}{x^2+1} + x \frac{2\frac{x+1}{x-1}}{\left( \frac{x+1}{x-1}\right)^2+1} = \frac{2x}{x^2+1} + \frac{2x(x^2-1)}{2x^2+2} = x\)
Consider \(H(x) = \frac{1}{1-x}\) then notice that \(H(H(x)) = \frac{1}{1-\frac{1}{1-x}} = \frac{x-1}{x}\) and \(H^3(x) = \frac{\frac{1}{1-x}-1}{\frac{1}{1-x}} = 1-(1-x) = x\). So
So letting \(S(x)\) be the statement that \(h(x) + h(H(x)) = 1 - x - \frac{1}{1-x}\) we have
\begin{align*}
S(x): && h(x) + h(H(x)) &= 1 - x - H(x) \\
S(H(x)): && h(H(x)) + h(H^2(x)) &= 1 - H(x) - H^2(x) \\
S(H^2(x)): && h(H^2(x)) + h(x) &= 1 - H^2(x) - x \\
S(x) - S(H(x)) + S(H^2(x)): && 2h(x) &= 1 - 2x \\
\Rightarrow && h(x)& = \frac12 - x
\end{align*}
and notice that \(\frac12 -x +\frac12 - \frac{1}{1-x} = 1 - x - \frac{1}{1-x}\) so it does satisfy the equation.