Official solution
f(x) = (3x -- 7)x2/3
f'(x) = \[\left( {3x - 7} \right){2 \over {3{x^{1/3}}}} + {x^{{2 \over 3}}}.3\]
= \[{{6x - 14 - 9x} \over {3{x^{1/3}}}}\]
= \[{{15x - 14} \over {3{x^{1/3}}}}\]
As f(x) increasing so f'(x) > 0
\[ \therefore \] \[{{15x - 14} \over {3{x^{1/3}}}}\] > 0
\[ \therefore \] x \[ \in \] \[\left( { - \infty ,0} \right) \cup \left( {{{14} \over {15}},\infty } \right)\]