If the function $f(x) = x(x+3)e^{-x/2}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$,then a root of $f'(x) = 0$ is

If the function $f(x) = x(x + 3) e^{-x/2}$ satisfies Rolle's theorem in the interval $[-3, 0]$,then find the value of $c$.

If $27a + 9b + 3c + d = 0$,then the equation $4ax^3 + 3bx^2 + 2cx + d = 0$ has at least one root between which of the following?

Consider the following functions:
$I) f(x) = \begin{cases} \frac{1}{2}-x, & x < \frac{1}{2} \\ (\frac{1}{2}-x)^2, & x \geq \frac{1}{2} \end{cases}$
$II) f(x) = |3x-1|$
$III) f(x) = x|x|$
$IV) f(x) = |x|$
Then on $[0, 1]$,Lagrange's Mean Value Theorem $(LMVT)$ is applicable to which of the functions?

Let $f(x)$ satisfy all the conditions of the Mean Value Theorem in $[0, 2]$. If $f(0) = 0$ and $|f'(x)| \le \frac{1}{2}$ for all $x$ in $[0, 2]$,then:

If $f(x) = \log(\sin x)$,$x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$,then the value of $c$ by applying Lagrange's Mean Value Theorem $(LMVT)$ is: