一条曲线的方程为 $y = 2x^2 - \frac{1}{2x} + 3$。

(a) 求驻点的坐标。

(b) 判断该驻点的性质。

(c) 对于正的 $x$ 值，判断这条曲线所表示的函数是递增、递减，还是两者都不是。请给出理由。

对分式项用负指数求导。令导数为零后，可以得到关于 $x^3$ 的方程。

For $x\ne0$, $y'=4x+\frac1{2x^2}$. At a stationary point,

Thus $x=-\frac12$, where $y=2(1/4)+1+3=\frac92$. The stationary point is $(-\frac12,\frac92)$.

计算二阶导数在驻点处的符号。

The second derivative is $y''=4-x^{-3}$. At $x=-\frac12$ it is $4-(-8)=12>0$, so the stationary point is a minimum.

把一阶导数通分。分母恒正，分子的立方函数在驻点两侧分别为负和正，因此函数先降后升。

For $x\ne0$,

The denominator is positive. For $x<-1/2$, the numerator is negative, so $y'<0$. For $-1/2<x<0$, it is positive, so $y'>0$. Thus the derivative changes from negative to positive at $x=-1/2$, and the stationary point is a minimum.

对正的 $x$ 分别检查一阶导数的两项。

For every $x>0$, both $4x$ and $1/(2x^2)$ are positive. Thus $y'>0$ on this interval, so the function is strictly increasing there.