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

Consider the function $f(x) = |x - 2| + |x - 5|, x \in R$.
Statement-$1$: $f'(4) = 0$.
Statement-$2$: $f$ is continuous in $[2, 5]$,differentiable in $(2, 5)$ and $f(2) = f(5)$.

In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$ and $f(x) = \sqrt{x}$,then the value of $c$ is:

If $f: R \rightarrow R$ is a twice differentiable function such that $f^{\prime \prime}(x) > 0$ for all $x \in R$,and $f(\frac{1}{2}) = \frac{1}{2}$,$f(1) = 1$,then

If $f(x) = x^{3}$ and $g(x) = x^{3} - 4x$ in the interval $[-2, 2]$,consider the following statements:
$(a)$ $f(x)$ and $g(x)$ satisfy the Mean Value Theorem.
$(b)$ $f(x)$ and $g(x)$ both satisfy Rolle's Theorem.
$(c)$ Only $g(x)$ satisfies Rolle's Theorem.
Which of these statements is correct?

If $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \end{cases}$, then Rolle's theorem is not applicable to $f(x)$ because