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

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 x = 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.