求函数U f(x)=2x^2- lnx的单调区间和极值.

由题意得：定义域$(0,+\infty )$

$f'(x)=4x-\dfrac {1} {x}=\dfrac {4{x}^{2}-1} {x}=\dfrac {(2x+1)(2x-1)} {x}$

令$f'(x)=\dfrac {(2x+1)(2x-1)} {x}\gt 0$

解得$x\gt \dfrac {1} {2}$或$x\lt -\dfrac {1} {2}$（舍去）

$\therefore $单调递增区间为$(\dfrac {1} {2},+\infty )$，单调递减区间为$(0,\dfrac {1} {2})$

当$x=\dfrac {1} {2}$时，取极小值$f(\dfrac {1} {2})=2\times \dfrac {1} {4}-ln\dfrac {1} {2}=\dfrac {1} {2}+ln2$

综上所述，结论为：单调递增区间为$(\dfrac {1} {2},+\infty )$，单调递减区间为$(0,\dfrac {1} {2})$；极小值$f(\dfrac {1} {2})=\dfrac {1} {2}+ln2$