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 $c$.

Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1, 0$ and $2$ be as given in the following table:

$x$	$x=-1, 0, 2$
$f(x)$	$3, 6, 0$
$g(x)$	$0, 1, -1$

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3g)^{\prime \prime}$ never vanishes. Then the correct statement$(s)$ is(are):
$(A)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup (0,2)$
$(B)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

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

If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

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

Let $f:[a, b] \rightarrow R$ be such that $f$ is differentiable in $(a, b)$, continuous at $x=a$ and $x=b$, and $f(a)=0=f(b)$. Then: