Q.Explain why Lagrange's mean value theorem is not applicable to the following functions in the respective intervals:

Verifies Rolle's hypotheses then solves $f'(x)=0$ for the point strictly inside $(0,3)$.

1. $f(x)=x^3-3x^2$ is a polynomial, so it is continuous on $[0,3]$ and differentiable on $(0,3)$.
2. $f(0)=0^3-3(0)^2=0$ and $f(3)=27-27=0$, so $f(0)=f(3)$ — all three hypotheses of Rolle's theorem hold.
3. Rolle's theorem guarantees at least one $c\in(0,3)$ with $f'(c)=0$.
4. $f'(x)=3x^2-6x=3x(x-2)$. Setting $f'(x)=0$: $x=0$ or $x=2$.
5. Of these, $x=0$ is an endpoint (excluded from the open interval), so the required value is $c=2$.

Checking $f(0)=f(3)$ and solving $f'(x)=0$ for the point guaranteed by Rolle's theorem inside $(0,3)$.

1. $f(x)=x^3-3x^2$ on $[0,3]$. $f(0)=0$, $f(3)=27-27=0$, so $f(0)=f(3)$ and Rolle's theorem applies (polynomial, so continuous and differentiable everywhere).
2. Rolle's theorem guarantees a $c\in(0,3)$ with $f'(c)=0$.
3. $f'(x)=3x^2-6x=3x(x-2)$. Setting $=0$: $x=0$ or $x=2$.
4. Only $x=2$ lies in the open interval $(0,3)$ (since $x=0$ is an endpoint).

Lagrange's Mean Value Theorem is the special case of Cauchy's Generalised Mean Value Theorem when $g(x)=x$ (the identity function).

1. Cauchy's Generalised Mean Value Theorem states: if $f,g$ are continuous on $[a,b]$, differentiable on $(a,b)$, and $g'(x)\neq0$, then there exists $c\in(a,b)$ with $\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$.
2. Lagrange's Mean Value Theorem states: there exists $c\in(a,b)$ with $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.
3. To recover Lagrange's form from Cauchy's form, we need $g(b)-g(a)=b-a$ and $g'(c)=1$.
4. Choosing $g(x)=x$ (the identity function) gives $g(b)-g(a)=b-a$ and $g'(x)=1$ for all $x$, which exactly reduces Cauchy's result to Lagrange's.
5. Hence Lagrange's law of mean is the particular case of the Generalised law of mean when $g(x)$ is an identity function.

Applying Lagrange's Mean Value Theorem to $f(x)=\sqrt x$ on $[1,4]$ and solving $f'(c)$ equal to the average rate of change gives $c=\dfrac94$.

1. LMVT states: for $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $c\in(a,b)$ with $f'(c) = \dfrac{f(b)-f(a)}{b-a}$.
2. Here $f(x)=\sqrt x$, $a=1$, $b=4$. Compute $f(1)=\sqrt1=1$ and $f(4)=\sqrt4=2$.
3. The average rate of change is $\dfrac{f(4)-f(1)}{4-1} = \dfrac{2-1}{3} = \dfrac13$.
4. Differentiate: $f'(x) = \dfrac{1}{2\sqrt x}$, so $f'(c) = \dfrac{1}{2\sqrt c}$.
5. Set $\dfrac{1}{2\sqrt c} = \dfrac13$, giving $2\sqrt c = 3$, i.e. $\sqrt c = \dfrac32$.
6. Squaring, $c = \dfrac94 = 2.25$, which indeed lies in $(1,4)$.

Verify Rolle's hypotheses on $[\pi,3\pi]$ and solve $f'(c)=0$; the only root of $\sin(c/2)=0$ lying strictly between $\pi$ and $3\pi$ is $c=2\pi$.

1. $f(x)=\cos\dfrac{x}{2}$ is continuous on $[\pi,3\pi]$ and differentiable on $(\pi,3\pi)$ (cosine is differentiable everywhere), so Rolle's theorem applies provided the end values match.
2. Check end values: $f(\pi)=\cos\dfrac{\pi}{2}=0$ and $f(3\pi)=\cos\dfrac{3\pi}{2}=0$. So $f(\pi)=f(3\pi)$, and Rolle's theorem guarantees at least one $c\in(\pi,3\pi)$ with $f'(c)=0$.
3. Differentiate: $f'(x) = -\dfrac{1}{2}\sin\dfrac{x}{2}$.
4. Set $f'(c)=0$: $-\dfrac12\sin\dfrac{c}{2}=0 \Rightarrow \sin\dfrac{c}{2}=0 \Rightarrow \dfrac{c}{2}=n\pi \Rightarrow c=2n\pi,\ n\in\mathbb{Z}$.
5. Among values $c=0,2\pi,4\pi,\dots$, the one lying strictly inside $(\pi,3\pi)$ is $c=2\pi$ (since $\pi<2\pi<3\pi$).
6. Hence $c=2\pi$, matching option (b).