Answer
$[0.5,0.75]$
Work Step by Step
The boundary values are when $I=3600$ and $I=1600$.
If $I=3600=\frac{900}{x^2}\\3600x^2=900\\x^2=0.25\\x=\pm0.5$
But $x$ is positive, thus $x=0.5$
If $I=1600=\frac{900}{x^2}\\1600x^2=900\\x^2=9/16\\x=\pm0.75$
But $x$ is positive, thus $x=0.75$
So the range is $[0.5,0.75]$
