1. ## Self-Inverse

$f_a (x) = \frac{2}{x } + \frac{x}{ 2} - a$ So i have done $f(f_a(x)) = x$ and got $\frac{2}{ \frac{2}{x } + \frac{x}{2 } -a } + \frac{\frac{2}{x }+\frac{x}{ 2} -a }{2 } -a = x$ I have then got simplified $\frac{5}{x } - \frac{2}{ a} + \frac{x}{4 } - \frac{3a}{2 } = 0$ and am now stuck, any help please? Thanks.

$f_a (x) = \frac{2}{x } + \frac{x}{ 2} - a$ So i have done $f(f_a(x)) = x$ and got $\frac{2}{ \frac{2}{x } + \frac{x}{2 } -a } + \frac{\frac{2}{x }+\frac{x}{ 2} -a }{2 } -a = x$ I have then got simplified $\frac{5}{x } - \frac{2}{ a} + \frac{x}{4 } - \frac{3a}{2 }$ and am now stuck, any help please? Thanks.
I'm a bit confused. Are you trying to find the value of a such that f_a is it's own inverse? And I'm not sure how you got your final equation.

If you are trying to find an a such that f_a is its own inverse:
$\frac{2}{ \frac{2}{x} + \frac{x}{2} - a } + \frac{\frac{2}{x}+\frac{x}{2} - a }{2} - a = x$

Clear the complex fractions:
$\frac{2}{ \frac{2}{x} + \frac{x}{2} - a } \cdot \frac{2x}{2x} + \frac{\frac{2}{x}+\frac{x}{2} - a }{2} \cdot \frac{2x}{2x} - a = x$

$\frac{4x}{4 + x^2 - 2ax} + \frac{4 + x^2 - 2ax}{4} - a = x$

Now put the whole equation under a common denominator, cancel the denominator, and this leaves you a polynomial in powers of x. You will see that the x^4 coefficient cannot be set to 0, so this problem cannot be solved for any value of a.

-Dan

3. I have got to find the 2-cycle $p_1(a), p_2(a)$ and its intervals of existence and stability. The way we have been shown in class in $f(f(x)) = x$ The answer is a $a \pm \sqrt{{a}^{ 2} - \frac{4}{ 3} }$. But from my answer i can not get this answer, so i think ive done something wrong

[snip] I have then got simplified $\frac{5}{x } - \frac{2}{ a} + \frac{x}{4 } - \frac{3a}{2 }$ and am now stuck, any help please? Thanks.