The maximum value of $\frac{\log x}{x}$ is

Let $f(x) = \begin{cases} 3 - x, & 0 \le x < 1 \\ x^2 + \log_e b, & x \ge 1 \end{cases}$. The set of values of $b$ such that $f(x)$ has a local minimum at $x = 1$ is

For the curve $y = x e^x$,the point:

On the interval $[1, e]$,the greatest value of $f(x) = x^2 \log x$ is:

If $f(x) = x + \frac{1}{x}$,$x \neq 0$,then the local maximum and minimum values of the function $f$ are respectively....

Let a function $f(x)$ be continuous in an interval $[a, b]$. Let $\delta > 0$ be a very small real number. Let $c \in (a, b)$ be such that $f(c - \delta) < f(c)$ and $f(c + \delta) < f(c)$ for every $\delta > 0$. Let $(f(\alpha - \delta) - f(\alpha))(f(\alpha + \delta) - f(\alpha)) < 0$ for all $\alpha \in (a, b)$ and $\alpha \neq c$. Then: