The function $f(x) = \frac{{2{x^2} - 1}}{{{x^4}}}$, (where, $\left. {x > 0} \right)$ decreases in the interval……………..

The function $f(x) = \frac{{2{x^2} - 1}}{{{x^4}}}$, where $x > 0$, decreases in the interval $(1,\infty )$.

${f^\prime }(x) = \frac{{{x^4} \cdot 4x - \left( {2{x^2} - 1} \right) \cdot 4{x^3}}}{{{x^8}}} = \frac{{4{x^5} - 8{x^5} + 4{x^3}}}{{{x^8}}}$

$= \frac{{ - 4{x^5} + 4{x^3}}}{{{x^8}}} = \frac{{4{x^3}\left( { - {x^2} + 1} \right)}}{{{x^8}}}$

Also, ${f^\prime }(x) < 0$

$\Rightarrow$ $\frac{{4{x^3}\left( {1 - {x^2}} \right)}}{{{x^8}}} < 0 \Rightarrow {x^2} > 1$

$\Rightarrow$ $x > \pm 1$

Therefore, $x \in (1,\infty )$