Maximum value of the function $f(x) = [x(x-1) + 1]^{\frac{1}{3}}$ for $0 \leq x \leq 1$ is . . . . . . .

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:

Find all the points of local maxima and local minima of the function $f$ given by $f(x) = 2x^3 - 6x^2 + 6x + 5$.

The abscissae of the points,where the tangent to the curve $y = x^3 - 3x^2 - 9x + 5$ is parallel to the $x$-axis,are

The maximum volume of the right circular cone with slant height $6$ units is

Let $M$ and $m$ be respectively the local maximum and the local minimum values of the function $f(x) = 2x^3 - 9x^2 + 12x + 5$ in the interval $[0, 3]$. Then $M - m$ is equal to