The value of $c$ satisfying the conditions and conclusions of Rolle's theorem for the function $f(x) = x \sqrt{x+6}$ on the interval $x \in [-6, 0]$ is:

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:

Given $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$,$g(x) = \begin{cases} \frac{\tan([x])}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$,$h(x) = \{x\}$,and $k(x) = 5^{\log_2(x + 3)}$. Then,in the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to:

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

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

If $f(x) = x^{\alpha} \log x, x > 0, f(0) = 0$ and $f(x)$ satisfies Rolle's theorem on $[0, 1]$,then what is the value of $\alpha$?