For all $x \in [0, 2024]$,assume that $f(x)$ is differentiable,$f(0) = -2$,and $f^{\prime}(x) \geq 5$. Then the least possible value of $f(2024)$ is:

Let $f(x)$ be continuous on $[0,4]$,differentiable on $(0,4)$,$f(0)=4$ and $f(4)=-2$. If $g(x)=\frac{f(x)}{x+2}$,then the value of $g^{\prime}(c)$ for some Lagrange's constant $c \in (0,4)$ is

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

Given that $f(x)$ is continuously differentiable on $a \le x \le b$ where $a < b, f(a) < 0$ and $f(b) > 0$,which of the following are always true?
$(i)$ $f(x)$ is bounded on $a \le x \le b$.
$(ii)$ The equation $f(x) = 0$ has at least one solution in $a < x < b$.
$(iii)$ The maximum and minimum values of $f(x)$ on $a \le x \le b$ occur at points where $f'(c) = 0$.
$(iv)$ There is at least one point $c$ with $a < c < b$ where $f'(c) > 0$.
$(v)$ There is at least one point $d$ with $a < d < b$ where $f'(d) < 0$.

In the Mean Value Theorem,$f'(c) = \frac{f(b) - f(a)}{b - a}$. If $a = 0$,$b = \frac{1}{2}$,and $f(x) = x(x - 1)(x - 2)$,then the value of $c$ is:

Suppose $f$ is differentiable for all $x$. If $f(1) = -2$ and $f'(x) \geq 2$ for all $x \in [1, 6]$,then: