Let $f(x) = \begin{cases} x^{3/5} & \text{if } x \le 1 \\ -(x - 2)^3 & \text{if } x > 1 \end{cases}$. Then the number of critical points on the graph of the function is:

$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given below:
$f'(x)$ sign chart:
- For $x < -5$,$f'(x) > 0$
- For $-5 < x < 2$,$f'(x) < 0$
- For $2 < x < 4$,$f'(x) > 0$
- For $x > 4$,$f'(x) < 0$
From the possible graph of $y = f(x)$,we can say that:

Let $h(x) = \min \{ x, x^2 \}$ for every real number $x$. Then:

If $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}$,then:

Let $f(x)$ be a quadratic expression which is positive for all real $x$. If $g(x) = f(x) + f'(x) + f''(x)$,then for any real $x$,which one is correct?

Let $f(x)$ be a non-negative differentiable function on $[0, \infty)$ such that $f(0)=0$ and $f^{\prime}(x) \leq 2 f(x)$ for all $x>0$. Then,on $[0, \infty)$: