Official solution
Let f(x) = \[{1 \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}\]
f'(x) = \[ - {1 \over 2}{{ - 6{x^2} - 18x + 12} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}\]
= \[{{ - 6\left( {x - 1} \right)\left( {x - 2} \right)} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}\]
fmax = f(1)
fmin = f(2)
f(1) = \[{1 \over {\sqrt {2 - 9 + 12 + 4} }}\] = \[{1 \over {\sqrt 9 }}\] = \[{1 \over 3}\]
f(2) = \[{1 \over {\sqrt {16 - 36 + 24 + 4} }}\] = \[{1 \over {\sqrt 8 }}\]
\[ \therefore \] \[{1 \over 3} < {I} <\] \[{1 \over {\sqrt 8 }}\]
So \[{1 \over 9} < {I^2} < {1 \over 8}\]