For $a>1$ we have:
$f:[\frac{1}{a},a]\to [\frac{1}{a},a]$ be a bijective function. Suppose
$f^{-1}(x)=\frac{1}{f(x)}$ for all $x \in [\frac{1}{a},a]$ then find $f(1)$.
Could someone give me slight hint for this question?
Also as function is bijective, hence it will be monotonic but can we tell for sure if it will be increasing or decreasing in this particular question?
From $f^{-1}(x)=\frac{1}{f(x)} \tag 1$
by replacing $x: = f(1)$ we get:
$1 = \frac{1}{f(f(1))}$ so $f(f(1))= 1$
By applying $f$ to (1) we get $x=f(\frac{1}{f(x)})$, so $1=f(\frac{1}{f(1)})$.
Therefore $f(f(1))=f(\frac{1}{f(1)})$ and, using injectivity, $f(1)=\frac{1}{f(1)}$. If follows $f(1)=1$.
About your claim "function is bijective, hence it will be monotonic", well, this is guarantee when $f$ continuos.