Let R be the set of all real numbers and f:[- | vedclass

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(D) Given, $f(x) = \begin{cases} x \sin \frac{1}{x}, & x 
eq 0 \ 0, & x=0 \end{cases}$.Continuity at $x=0$: $LHL = \lim_{x ightarrow 0^-} x \sin \

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$f$ satisfies the conditions of Lagrange's mean value theorem on $[0,1]$

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(D) Given, $f(x) = \begin{cases} x \sin \frac{1}{x}, & x 
eq 0 \ 0, & x=0 \end{cases}$.Continuity at $x=0$: $LHL = \lim_{x ightarrow 0^-} x \sin \frac{1}{x} = 0$ and $RHL = \lim_{x ightarrow 0^+} x \sin \frac{1}{x} = 0$. Since $LHL = RHL = f(0)$, $f(x)$ is continuous for all $x \in [-1, 1]$.Differentiability at $x=0$: $f'(0) = \lim_{h ightarrow 0} \frac{f(h) - f(0)}{h} = \lim_{h ightarrow 0} \frac{h \sin(1/h)}{h} = \lim_{h ightarrow 0} \sin(1/h)$. This limit does not exist as it oscillates between $-1$ and $1$. Thus, $f(x)$ is not differentiable at $x=0$.Rolle's theorem requires $f(a) = f(b)$ and differentiability on $(a, b)$. Since $f(x)$ is not differentiable at $x=0$, it does not satisfy Rolle's theorem on $[-1, 1]$ or $[0, 1]$.Lagrange's Mean Value Theorem $(LMVT)$ requires continuity on $[a, b]$ and differentiability on $(a, b)$. Since $f(x)$ is not differentiable at $x=0$, it does not satisfy $LMVT$ on $[-1, 1]$ (as $0 \in (-1, 1)$) or $[0, 1]$ (as $0$ is an endpoint, but the theorem requires differentiability on the open interval $(0, 1)$).Wait, checking the options: The function is differentiable on $(0, 1)$. Thus, $f(x)$ satisfies the conditions of $LMVT$ on $[0, 1]$ because it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.

Let $R$ be the set of all real numbers and $f:[-1,1] \rightarrow R$ be defined as $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}$. Then:

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