 # Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals `
Question:

Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals

(i) $f(x)=3+(x-2)^{2 / 3}$ on $[1,3]$

(ii) $f(x)=[x]$ for $-1 \leq x \leq 1$, where $[x]$ denotes the greatest integer not exceeding $x$

(iii) $f(x)=\sin \frac{1}{x}$ for $-1 \leq x \leq 1$

(iv) $f(x)=2 x^{2}-5 x+3$ on $[1,3]$

(v) $f(x)=x^{2 / 3}$ on $[-1,1]$

(vi) $f(x)= \begin{cases}-4 x+5, & 0 \leq x \leq 1 \\ 2 x-3, & 1 Solution: (i) The given function is$f(x)=3+(x-2)^{\frac{2}{3}}$. Differentiating with respect to x, we get$f^{\prime}(x)=\frac{2}{3}(x-2)^{\frac{2}{3}-1}\Rightarrow f^{\prime}(x)=\frac{2}{3}(x-2)^{\frac{-1}{3}}\Rightarrow f^{\prime}(x)=\frac{2}{3(x-2)^{\frac{1}{3}}}$Clearly, we observe that for$x=2 \in[1,3], f^{\prime}(x)$does not exist. Therefore,$f(x)$is not derivable on$[1,3]$. Hence, Rolle's theorem is not applicable for the given function. (ii) The given function is$f(x)=[x]$. The domain of$f$is given to be$[-1,1]$. Let$c \in[-1,1]$such that$c$is not an integer. Then,$\lim _{x \rightarrow c} f(x)=f(c)$Thus,$f(x)$is continuous at$x=c$. Now, let$c=0$Then,$\lim _{x \rightarrow 0^{-}} f(x)=-1 \neq 0=f(0)$Thus,$f$is discontinuous at$x=0$Therefore,$f(x)$is not continuous in$[-1,1]$. Rolle's theorem is not applicable for the given function. (iii) The given function is$f(x)=\sin \frac{1}{x}$. The domain of$f$is given to be$[-1,1]$. It is known that$\lim _{x \rightarrow 0} \sin \frac{1}{x}$does not exist. Thus,$f(x)$is discontinuous at$x=0$on$[-1,1]$. Hence, Rolle's theorem is not applicable for the given function. (iv) The given function is$f(x)=2 x^{2}-5 x+3$on$[1,3]$. The domain of$f$is given to be$[1,3]$. It is a polynomial function. Thus, it is everywhere derivable and hence continuous. But$f(1)=0$and$f(3)=6\Rightarrow f(3) \neq f(1)$Hence, Rolle's theorem is not applicable for the given function. (v) The given function is$f(x)=x^{\frac{2}{3}}$on$[-1,1]$. The domain of$f$is given to be$[-1,1]$. Differentiating$f(x)$with respect to$\mathrm{x}$, we get$f^{\prime}(x)=\frac{2}{3} x^{-\frac{1}{3}}$We observe that at$x=0, f^{\prime}(x)$is not defined. Hence, Rolle's theorem is not applicable for the given function. (vi) The given function is$f(x)=\left\{\begin{array}{l}-4 x+5,0 \leq x \leq 1 \\ 2 x-3,1

At = 0, we have

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[-4(1-h)+5]=1$

And

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[2(1+h)-3]=-1$

$\therefore \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$

Thus, $f(x)$ is discontinuous at $x=1$.

Hence, Rolle's theorem is not applicable for the given function.