Not A Paper Pen Problem!

Let $f (x)=\frac {x^{3}+1}{2}$ . Then the equation rearranges to $f (f (x))=x$ .

Note that $f (x)$ is a strictly increasing function meaning that $a <b \implies f (a)<f (b)$ where $a, b$ are reals.

Thus,

$f (x)> x \implies f (f (x))> f (x)> x, f (x)<x \implies f (f (x))<f (x)<x$

For equality to occur, $f (x)=x \implies \frac {x^{3}+1}{2}=x \implies x^{3}-2x+1=(x-1)(x^{2}+x-1)=0$

This gives the solutions $1, \frac {-1 \pm \sqrt {5}}{2}$ . The only negative solution is $\frac {-1-\sqrt {5}}{2}$ .

The equation can be rewritten as: $\frac { { x }^{ 3 }+1 }{ 2 } =\sqrt [ 3 ]{ 2x-1 }$ .Now notice that R.H.S. is the $inverse$ of L.H.S. or vice versa.Hence,solution occurs for $y=x$ .Therefore, $\Rightarrow \frac { { x }^{ 3 }+1 }{ 2 } =x\\ \Rightarrow { x }^{ 3 }-2x+1=0$ .On solving we get, $x=1,\frac { -1\pm \sqrt { 5 } }{ 2 }$ .Hence only negative solution is $x=\frac { -1-\sqrt { 5 } }{ 2 } \simeq \boxed { -1.61 }$

'Cause I didn't realize RHS and LHS are inverses, here's my way of solving:

Let $y=\sqrt[3]{2x-1}$ , then $y^3=2x-1$ , or $y^3+1=2x$ .

The equation now changes to this system: $\begin{cases}x^3+1=2y\\y^3+1=2x \end{cases}$ .

Subtract the first and the second equations, we have: $(x-y)(x^2+xy+y^2)=2(y-x)$ (1)

If $x \neq y$ , (1) $\Leftrightarrow x^2+xy+y^2=-2 \Leftrightarrow (x+\frac{1}{2}y)^2+\frac{3}{4}y^2=-2$ (false for all $x,y$ ).

So, $x=y$ , given $x^3+1=2x$ . There are 3 solutions for this equation, but the only negative one is $\frac{-1-\sqrt{5}}{2}$ .

by considering the inverse of x on both sides i found that x<0 )so i used (x=-1) i solved for x= on one side and found that divergence occurs towards a root or is that not allowed? im new to this brilliant. org thing