I want to show that a function has a fixed point in a given interval. Here is the setup:
Let $f : [1,3] \mapsto \mathbb{R}$ be defined by $f(x) = 3^{g(x)}$ with $g(x) = \frac{3}{2} - |x - \frac{x^2}{2}| $. Show that $f$ has a fixed point in $[1,3]$
I know that the fixed point is where $f(x) = x$ (where the graphs cross each other)
but we are not allowed to use a calculator so the calculation becomes problematic (Otherwise I would use $ln$). Am I missing something crucial here e.g. could I simplify the function somehow? I don't know if it would even be useful here but we are not allowed to use derivatives.
Define $$h(x) = x - f(x).$$ Because $f$ and $x$ are continuous so is $h$. Now recall the Intermediate Value Theorem which is saying if a function $h$ is continuous on an interval $[a,b]$ and if $h(a)<0$ and $h(b)>0$ , then there is some $c $ with $a<c<b$ so that $h(c)=0. $ By direct calculations we get $$h(2)=2-f(2)= 2- 3^{g(2)} = 2 - 3^{1.5}<0 $$ and $$h(3)= 2- 3^{g(3)}= 2-3^0= 1 >0.$$ Thus there exists $ x$ between $2$ and $3$ such that $h(x)=0$. Therefore $h(x)=x-f(x)=0$. This means $f(x) =x$.